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Instructor’s Resource Manua 

to accompany 



Electronic Devices and 
Circuit Theory 



Tenth Edition 



Robert L. Boylestad 
Louis Nashelsky 




Upper Saddle River, New Jersey 
Columbus, Ohio 



www.elsolucionario.net 





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Copyright © 2009 by Pearson Education, Inc., Upper Saddle River, New Jersey 07458. 

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Instructors of classes using Boylestad/Nashelsky, Electronic Devices and Circuit Theory , 10 th edition, may 
reproduce material from the instructor’s text solutions manual for classroom use. 



10 987654321 



PEARSON 



ISBN-13: 978-0-13-503865-9 
ISBN-10: 0-13-503865-0 




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Contents 



Solutions to Problems in Text 1 



Solutions for Laboratory Manual 185 



iii 



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Chapter 1 



1. Copper has 20 orbiting electrons with only one electron in the outermost shell. The fact that 
the outermost shell with its 29 th electron is incomplete (subshell can contain 2 electrons) and 
distant from the nucleus reveals that this electron is loosely bound to its parent atom. The 
application of an external electric field of the correct polarity can easily draw this loosely 
bound electron from its atomic structure for conduction. 

Both intrinsic silicon and germanium have complete outer shells due to the sharing (covalent 
bonding) of electrons between atoms. Electrons that are part of a complete shell structure 
require increased levels of applied attractive forces to be removed from their parent atom. 

2. Intrinsic material: an intrinsic semiconductor is one that has been refined to be as pure as 
physically possible. That is, one with the fewest possible number of impurities. 

Negative temperature coefficient: materials with negative temperature coefficients have 
decreasing resistance levels as the temperature increases. 

Covalent bonding: covalent bonding is the sharing of electrons between neighboring atoms to 
form complete outermost shells and a more stable lattice structure. 



4. W= QV= (6 C)(3 V) = 18 J 



5. 



48 eV = 48(1.6 x 1(T 19 J) = 76.8 x 10 19 



Q= W_ = 76.8x10" 
^ V 12 V 



— = 6.40 x 10 19 



J 

C 



6.4 x 10 19 C is the charge associated with 4 electrons. 



6. GaP Gallium Phosphide 
ZnS Zinc Sulfide 



E g = 2.24 eV 
E g = 3.67 eV 



7. An / 2 -type semiconductor material has an excess of electrons for conduction established by 
doping an intrinsic material with donor atoms having more valence electrons than needed to 
establish the covalent bonding. The majority carrier is the electron while the minority carrier 
is the hole. 



A p - type semiconductor material is formed by doping an intrinsic material with acceptor 
atoms having an insufficient number of electrons in the valence shell to complete the covalent 
bonding thereby creating a hole in the covalent structure. The majority carrier is the hole 
while the minority carrier is the electron. 

8. A donor atom has five electrons in its outermost valence shell while an acceptor atom has 
only 3 electrons in the valence shell. 

9. Majority carriers are those carriers of a material that far exceed the number of any other 
carriers in the material. 

Minority carriers are those carriers of a material that are less in number than any other carrier 
of the material. 



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10 . 



Same basic appearance as Fig. 1.7 since arsenic also has 5 valence electrons (pentavalent). 
Same basic appearance as Fig. 1.9 since boron also has 3 valence electrons (trivalent). 



12 . 

13. 

14. For forward bias, the positive potential is applied to the p - type material and the negative 
potential to the n- type material. 

15. 7^ = 20 + 273 = 293 

k = 1 1,600 In = 1 1,600/2 (low value of V D ) = 5800 



f kV D > 




f (5800)(0.6) \ 


e lK - 1 


= 50 x 10“ 9 


e 293 - 1 






v ) 


) 







= 50 x 10“ 9 (e 1L877 - 1) = 7.197 mA 

16. k= 1 1,600/ft = 1 1,600/2 = 5800 (n = 2 for V D = 0.6 V) 
T k =T c + 273 = 100 + 273 = 373 

(5800)(0.6 V) 

e kv,T *=e 373 = e 9 ' 33 = 11.27 xlO 3 

7 = I s (e WiTK -1) = 5 juA( 1 1.27 x 10 3 - 1) = 56.35 mA 

17. (a) 7^=20 + 273=293 

k= 1 1,600/m = 1 1,600/2 = 5800 



Id = I, 


f kV D > 

e TK - 1 


= 0.1//A 


f (5800)(-10 V) \ 

e 293 - 1 




l J 




v / 



= o.i x io~V 19795 - 1) = 0.1 x 1 0 -6 ( 1 .07 x 10“ 86 - 1) 

= 0.1 x 10“ 6 0 . 1 / jA 

I D = I S = 0.1 

(b) The result is expected since the diode current under reverse-bias conditions should equal 
the saturation value. 

18. (a) 



X 


T = e X 


0 


1 


1 


2.7182 


2 


7.389 


3 


20.086 


4 


54.6 


5 


148.4 



(b) y = e°=l 

(c) For V= 0 V, e° = 1 and/ = 7/1 - 1) = 0 mA 



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19. T= 20°C: 4 = 0.1 juA 

T= 30°C: 4 = 2(0.1 //A) = 0.2 //A (Doubles every 10°C rise in temperature) 
T= 40°C: 4 = 2(0.2 //A) = 0.4 //A 
T= 50°C: 4 = 2(0.4 //A) = 0.8 juA 
T= 60°C: 4 = 2(0.8 //A) = 1.6 //A 

1.6 //A: 0.1 /jA =>16:1 increase due to rise in temperature of 40°C. 



20. For most applications the silicon diode is the device of choice due to its higher temperature 
capability. Ge typically has a working limit of about 85 degrees centigrade while Si can be 
used at temperatures approaching 200 degrees centigrade. Silicon diodes also have a higher 
current handling capability. Germanium diodes are the better device for some RF small signal 
applications, where the smaller threshold voltage may prove advantageous. 



21. From 1.19: 





-75°C 


25°C 


125°C 


V F 

@ 10 mA 


1.1 V 


0.85 V 


0.6 V 


4 


0.01 pA 


1 pA 


1.05 juA 



Vf decreased with increase in temperature 
1.1 V: 0.6 V = 1.83:1 
4 increased with increase in temperature 
1.05 juA: 0.01 pA= 105 xl0 3 :l 



22. An “ideal” device or system is one that has the characteristics we would prefer to have when 
using a device or system in a practical application. Usually, however, technology only 
permits a close replica of the desired characteristics. The “ideal” characteristics provide an 
excellent basis for comparison with the actual device characteristics permitting an estimate of 
how well the device or system will perform. On occasion, the “ideal” device or system can be 
assumed to obtain a good estimate of the overall response of the design. When assuming an 
“ideal” device or system there is no regard for component or manufacturing tolerances or any 
variation from device to device of a particular lot. 

23. In the forward-bias region the 0 V drop across the diode at any level of current results in a 
resistance level of zero ohms - the “on” state - conduction is established. In the reverse-bias 
region the zero current level at any reverse-bias voltage assures a very high resistance level - 
the open circuit or “off’ state - conduction is interrupted. 



24. The most important difference between the characteristics of a diode and a simple switch is 
that the switch, being mechanical, is capable of conducting current in either direction while 
the diode only allows charge to flow through the element in one direction (specifically the 
direction defined by the arrow of the symbol using conventional current flow). 



25 . 



V D = 0.66 V, I D = 2 mA 




0.65 V 
2mA 



= 325 Q 



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26 . 



At I D = 15 mA, V D = 0.S2 V 
0.82 V 
~~ I D 15 mA 

As the forward diode current increases, the static resistance decreases. 



27. V D = -10 V, I D = I s = -0.1 juA 



10 V 
0.1 juA 



= 100 MQ 



V D = -30 V, I D = I s = -0.1 juA 
V D _ 30 V 
I D O.ljuA 



= 300 MQ 



28. 



29. 



30. 



31. 



As the reverse voltage increases, the reverse resistance increases directly (since the diode 
leakage current remains constant). 



(a) r d =^f- 
A I, 



0.79 V -0.76 V 0.03 V 



15 mA -5 mA 10 mA 



= 3 Q 



26 mV 26 mV _ , ^ 

(b) r d = = = 2.6 Q 

I n 10 mA 



(c) quite close 

I D = 10 mA, V D = 0.76 V 



^DC ~ 



0.76 V 
I D 10 mA 



r d = 



AY, 0.79 V -0.76 V 0.03 V 



A L 



15 mA -5 mA 10 mA 



= 3 Q 



RdC >> 



j \ AV d 0.72 V -0.61V 
I D = 1 mA, r d = — 1 - = — — = 55 Q 



A /„ 



2 mA - 0 mA 



, ic a AV d 0.8 V-0.78 V 

//) = 1 5 mA, r d = — - = = 2 Q 

AI, 20 mA-10 mA 



Id — 1 mA, r d = 2 



^26 mV^ 



= 2(26 Q) = 52 Q vs 55 Q (#30) 



\ ± D J 



T ^ . 26 mV 26 mV „ ^ ^ ^ 

Id = 1 5 mA, r d = = = 1.73 Q vs 2 Q (#30) 

I n 15 mA 



32. 



f av 



AV d _ 0.9 V -0.6 V 

AI , 1 3.5 mA - 1 .2 mA 



= 24.4 Q 



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(c) 0.194 pF/V: 0.033 pF/V = 5.88:1 = 6:1 
Increased sensitivity near V D = 0 V 

37. From Fig. 1.33 

V D = 0 V, C D = 3.3 pF 
V D = 0.25 V, C D = 9 pF 

38. The transition capacitance is due to the depletion region acting like a dielectric in the reverse- 
bias region, while the diffusion capacitance is determined by the rate of charge injection into 
the region just outside the depletion boundaries of a forward-biased device. Both 
capacitances are present in both the reverse- and forward-bias directions, but the transition 
capacitance is the dominant effect for reverse-biased diodes and the diffusion capacitance is 
the dominant effect for forward-biased conditions. 



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39. 



V D = 0.2 V, C D = 7.3 pF 

X c = — = = 3.64 kQ 

2 nfC 2 tt{6 MHz)( 7.3 pF) 

V D = -20 V, C T = 0.9 pF 

X c = — = = 29.47 kQ 

2 nfC 2 tt(6 MHz)( 0.9 pF) 




42. As the magnitude of the reverse-bias potential increases, the capacitance drops rapidly from a 
level of about 5 pF with no bias. For reverse-bias potentials in excess of 10 V the capacitance 
levels off at about 1.5 pF. 

43. At V D = -25 V, I D = -0.2 nA and at V D = -100 V, I D = -0.45 nA. Although the change in I R is 
more than 100%, the level of I R and the resulting change is relatively small for most 
applications. 

44. Log scale: T A = 25°C, I R = 0.5 nA 

T a = 100°C, I R = 60 nA 
The change is significant. 

60 nA: 0.5 nA= 120:1 

Yes, at 95°C I R would increase to 64 nA starting with 0.5 nA (at 25°C) 

(and double the level every 10°C). 



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45. I F = 0.1 mA: r d =z 700 Q 
If — 1.5 mA: r d = 70 Q 
If = 20 mA: /;/ = 6Q 

The results support the fact that the dynamic or ac resistance decreases rapidly with 
increasing current levels. 



46. 



T= 25°C: P max = 500rnW 
T= 100°C: P max = 260 mW 

^ max = VfIf 




500 mW 
0.7 V 
260 mW 
0.7 V 



= 714.29 mA 
= 371.43 mA 



714.29 mA: 371.43 mA = 1.92:1 = 2:1 



47. Using the bottom right graph of Fig. 1 .37 : 
I F = 500 mA @T= 25°C 
At I F = 250 mA, T= 104°C 




0 75 V 

0.072= — — — — xlOO 

10 V(7] -25) 



7 5 

0.072 = 

7; -25 

u - 25° = = 104.17° 

0.072 

U = 104.17° + 25° = 129.17° 



50. 



Tr = 



AV 7 



x 100% 



V z (T x -T 0 ) 

(5 V - 4.8 V) 
5 V(100°-25°) 



x 100% = 0.053 %/°C 



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51. 



(20 V -6.8 V) 
(24 V -6.8 V) 



x 100% = 77% 



The 20 V Zener is therefore = 77% of the distance between 6.8 V and 24 V measured from 
the 6.8 V characteristic. 



At I z = 0.1 mA, T c = 0.06%/°C 
(5 V - 3.6 V) 



(6.8 V -3.6 V) 



x 100% = 44% 



The 5 V Zener is therefore = 44% of the distance between 3.6 V and 6.8 V measured from the 
3.6 V characteristic. 

At I z =0.1 mA, T c = -0.025%/°C 



52. 




24 V 



53. 



54. 



55. 



56. 



24 V Zener: 
0.2 mA 
1 mA 
10 mA 



= 400 Q 
= 95 0. 
= 13 Q 



The steeper the curve (higher dl/dV) the less the dynamic resistance. 

F T =2.0 V, which is considerably higher than germanium (= 0.3 V) or silicon (= 0.7 V). For 
germanium it is a 6.7:1 ratio, and for silicon a 2.86:1 ratio. 

Fig. 1.53 (f) I F =13 mA 
Fig. 1.53 (e) V F = 2.3 V 

(a) Relative efficiency @ 5 mA = 0.82 
@ 10 mA = 1.02 

1.02-0.82 



ratio: 



0.82 

1.02 



0.82 



x 100% = 24.4% increase 
= 1.24 



(b) Relative efficiency @30 mA = 1.38 
@ 35 mA = 1.42 

1 42 — 1 38 

x 100% = 2.9% increase 
ratio: = 1.03 



1.38 
1.42 



1.38 



(c) For currents greater than about 30 mA the percent increase is significantly less than for 
increasing currents of lesser magnitude. 



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57. (a) — = 0.25 

3.0 

From Fig. 1.53 (i) jC = 75° 

(b) 0.5 => jC = 40 ° 

58. For the high-efficiency red unit of Fig. 1.53: 





20 mA 


Average 


| \ ^-0.2mA/°C 


forward 


i 


current 






25 50 75 100 125 150 ' ' ° c 




100°C 



0.2 mA 20 mA 



°C x 

20 mA 
0.2 mA/°C 



100°C 



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Chapter 2 



The load line will intersect at I D = — = 



8 V 



R 330 Q 



= 24.24 mA and V n = 8 V. 



(a) V Dg = 0.92 V 
I n = 21.5 mA 

U Q 

Vr = E- V Dq = 8 V - 0.92 V = 7.08 V 

(b) V Dg = 0.7 V 

I n = 22.2 mA 

U Q 

Vr = E-V Dq = 8 V - 0.7 V = 7.3 V 

(c) F, e =0V 

A ^ 24.24 mA 

U Q 

Vr = E- V Dq =8 V-0 V = 8 V 

For (a) and (b), levels of V D and I D are quite close. Levels of part (c) are reasonably close 
but as expected due to level of applied voltage E. 



2 . 



(a) 




5 V 
2.2 kQ 



= 2.27 mA 



The load line extends from I D = 2.27 mA to V D = 5 V. 
= 0.7 V, I n = 2 mA 

Uq 1 Uq 



(b) 



I D = — = 5 V = 10.64mA 

0.47 kQ 

The load line extends from I D = 10.64 mA to V D = 5 V. 
V Dq - 0.8 V, I Dq = 9 mA 



(c) = 



5 V 



= 27.78 mA 



7? 0.18 kQ 

The load line extends from I D = 27.78 mA to V D = 5 V. 

= 0.93 V, / D = 22.5 mA 



The resulting values of V D are quite close, while I D extends from 2 mA to 22.5 mA. 



3. Load line through I D =10 mA of characteristics and V D = 7 V will intersect I D axis as 
11.25 mA. 

1 D = 11.25 mA= — = — 

R R 

with R = — — — = 0.62 kQ 
1 1.25 mA 



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4 . 



, , r r E-V d 30V-0.7V A 

(a) I D = I R = — = = 13.32 mA 

R 2.2 kQ 

V D = 0.7 V,V r = E- V d = 30 V- 0.7 V = 29.3 V 



r E-V d 30V-0V 4 

(b) I D = = = 13.64 mA 

R 2.2 kQ 

V D = 0 V, V R = 30 V 

Yes, since E V T the levels of I D and 17? are quite close. 

(a) 7=0 mA; diode reverse-biased. 

(b) F 20 q = 20 V - 0.7 V = 19.3 V (Kirchhoff s voltage law) 

19 3 V 

7= - =0.965 A 



(c) 7 = 



20 Q 
10 V 
10Q 



= 1 A; center branch open 



(a) Diode forward-biased, 

Kirchhoff s voltage law (CW): -5 V + 0.7 V - V Q = 0 

V 0 = -4.3 V 

4.3 V 



Ir= In = 



R 2.2 kQ 



= 1.955 mA 



(b) Diode forward-biased, 

, _ 8V-0.7V 

Id ~ 



= 1.24 mA 



1.2 kQ + 4.7 kQ 
V 0 = V 4Jkn + V D = (1.24 mA)(4.7 kQ) + 0.7 V 

= 6.53 V 



7. 



(a) V 0 = 



2kQ(20 V -0.7 V -0.3 V) 
2 kQ + 2 kQ 



= -7(20 V- 1 V)= 7(19 V) = 9.5 V 

_ 10 V + 2V - 0.7 V) 11.3V iai , . 

(b) 7= - = = 1.915 mA 

1.2 kQ + 4.7 kQ 5.9 kQ 

V'=IR = (1.915 mA)(4.7 kQ) = 9 V 
V Q = F'-2 V = 9 V-2 V = 7 V 



11 



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8 . 



(a) Determine the Thevenin equivalent circuit for the 10 mA source and 2.2 kQ resistor. 



E Th =IR = (10 mA)(2.2 kQ) = 22 V 
R Th = 2. 2kQ 




(b) 



Diode forward-biased 



Id = 



20 V+5 V -0.7 V 
6.8 kQ 



= 2.65 mA 



Kirchhoff s voltage law (CW): 
+V Q - 0.7 V + 5 V = 0 

V 0 = -4.3 V 



Diode forward-biased 



22 V-0.7 V 
2.2 kQ + 1.2 kQ 



= 6.26 mA 



K = I d ( 1.2 kQ) 

= (6.26 mA)( 1.2 kQ) 

= 7.51 V 



9. 



(a) V Qi = 12 V-0.7 V= 11.3 V 

V = 0.3 V 

°2 

(b) V Qx =-10 V + 0.3 V + 0.7 V = -9 V 

T 10V-0.7V-0.3V 9 V 

/ = = = 2 mi 

1.2 kQ + 3.3 kQ 4.5 kQ 



= -(2 mA)(3.3 kQ) = -6.6 V 



10. (a) Both diodes forward-biased 

20 V-0.7 V 



Ir = 



= 4.106 mA 



4.7 kQ 
Assuming identical diodes: 

/, _4J06mA _ M5 
2 2 

V 0 = 20 V - 0.7 V = 19.3 V 



(b) Right diode forward-biased: 

, _ 15 V + 5 V-0.7 V 

1 D 



2.2 kQ 

V 0 = 15 V - 0.7 V = 14.3 V 



= 8.77 mA 



11. (a) Ge diode “on” preventing Si diode from turning “on” 

T 10V-0.3V 9.7V 4 

/= = = 9.7 mA 



1 kQ 



1 kQ 



(b) 



16 V-0.7 V-0.7 V-12V _ 2.6 V 
4.7 kQ ~ 4.7 kQ 

V 0 = 12 V + (0.553 mA)(4.7 kQ) = 14.6 V 



= 0.553 mA 



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12 . 



Both diodes forward-biased: 

V. = 0.7 V, V. = 0.3 V 

°\ °2 



h kQ 



h 



20 V- 0.7 V 19.3 V 



1 kQ 

0.7 V- 0.3 V 



1 kQ 
= 0.851 mA 



= 19.3 mA 



0.47 kQ 
/(Si diode) = I\ m — /0.47 kQ 

= 19.3 mA -0.851 mA 

= 18.45 mA 



13. For the parallel Si - 2 kQ branches a Thevenin equivalent will result (for “on” diodes) in a 
single series branch of 0.7 V and 1 kQ resistor as shown below: 



10V 



0 — — °“ 



-° V, 



0.7 V 



1 kQ 



_ 2 kQ (10 V - 0.7 V) _ 2 f03v) 
1 kQ + 2 kQ 3 K 



E Th, R Th 



:2kQ = 6.2 V 



, 6.2 V . 

= -1.55 mA 



14. Both diodes “off’. The threshold voltage of 0.7 V is unavailable for either diode. 

Vo = 0V 

15. Both diodes “on”, V 0 = 10 V - 0.7 V = 9.3 V 

16. Both diodes “on”. 

V 0 = 0.7 V 

17. Both diodes “off’, V Q = 10 V 

18. The Si diode with -5 V at the cathode is “on” while the other is “off’. The result is 

V 0 = -5 V + 0.7 V = -4.3 V 

19. 0 V at one terminal is “more positive” than -5 V at the other input terminal. Therefore 
assume lower diode “on” and upper diode “off’. 

The result: 

Vo = 0 V - 0.7 V = -0.7 V 
The result supports the above assumptions. 

20. Since all the system terminals are at 10 V the required difference of 0.7 V across either diode 
cannot be established. Therefore, both diodes are “off’ and 

Vo = +10 V 

as established by 10 V supply connected to 1 kQ resistor. 



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21 . The Si diode requires more terminal voltage than the Ge diode to turn “on”. Therefore, with 
5 V at both input terminals, assume Si diode “off’ and Ge diode “on”. 

The result: V Q = 5 V - 0.3 V = 4.7 V 
The result supports the above assumptions. 




23. Using V dc = 0.3 18(F m -V T ) 

2 V = 0.318(T W - 0.7 V) 
Solving: V m = 6.98 V = 10:1 for V m :V T 






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W 2.2 k.Q) = = 2.855 mA 

2.2 kQ 

I D — I L + / max (2.2 kQ) = 0.924 mA + 2.855 mA = 3.78 mA 




25. V m = V2 (1 10 V) = 155.56 V 

V dc = 0.31 8 = 0.318(155.56 V) = 49.47 V 



aU, 



155.56 V 







26. 



Diode will conduct when v Q = 0.7 V; that is, 



v Q = 0.7 V = 



10 kQ(v.) 
lOkQ + lkQ 



Solving: v,- = 0.77 V 



For Vi > 0.77 V Si diode is “on” and v Q = 0.7 V. 

For Vi < 0.77 V Si diode is open and level of v 0 is determined 
by voltage divider rule: 



10 kQ(v.) 



lOkQ + lkQ 



= 0.909 v z 



For Vi = -10 V: 

v 0 = 0.909(-10 V) 

= -9.09 V 




When v 0 = 0.7 V, - 0.7 V 



/max(reverse) 



= 10 V-0.7 V = 9.3 V 



9.3 V 
1 kQ 



= 9.3 mA 



10 V 

1 kQ + 10kQ 



= 0.909 mA 




15 



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27. (a) P max =14 mW = (0.7 V)Id 

14 mW 



In = 



0.7 V 



= 20 mA 



(b) 4.7 kQ || 56 kQ = 4.34 kQ 
V R = 160 V - 0.7 V = 159.3 V 



7m ax 



159.3 V 
4.34 kQ 



= 36.71 mA 



(c) /diode =^Y= 36 ' 7 ^ mA = 18-36 mA 

(d) Yes, I D = 20 mA > 1 8.36 mA 

(d) Tdiode — 36.71 mA Ai a x = 20 mA 

28. (a) V m = V2 (120 V) = 169.7 V 

^ = V im -2 V d 

= 169.7 V- 2(0.7 V)= 169.7 V- 1.4 V 
= 168.3 V 

Vd C = 0.636(168.3 V) = 107.04 V 

(b) PIV = K m (load) + V D = 168.3 V + 0.7 V = 169 V 

,, , V L m 168.3 V A 

(c) //)(max) = — — = = 168.3 mA 

R l 1 kQ 

(d) P max = VrJ D = (0.7 V)/ max 

= (0.7 V)(l 68.3 mA) 

= 117.81 mW 




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30. 



Positive half-cycle of v{. 



Network redrawn: 




Negative half-cycle of : 



Voltage-divider rule: 

2.2 k Q(K ) 

y = w 

° max 2.2 kQ + 2.2 kQ 




= 1(100 V) 

= 50 V 







50 V 



Polarity of v 0 across the 2.2 kQ 
resistor acting as a load is the same. 

Voltage-divider rule: 

2.2kQ(K ) 

y = w 

° max 2.2 kQ + 2.2 kQ 




= f (100 V) 

= 50 V 



//\f F dc = 0.636K m = 0.636 (50 V) 

7 = 31.8 V 



3 1 . Positive pulse of v* : 

Top left diode “off”, bottom left diode “on” 
2.2 kQ || 2.2 kQ =1.1 kQ 

= kQ(l 70 V) = 56 67 v 

w 1.1 kQ + 2.2 kQ 



Negative pulse of v*: 

Top left diode “on”, bottom left diode “off’ 

= Llk Q (17Q = 56.67 V 
° peak 1.1 kQ + 2.2 kQ 

V dc = 0.636(56.67 V) = 36.04 V 

32. (a) Si diode open for positive pulse of v* and v 0 = 0 V 

For -20 V < Vi < -0.7 V diode “on” and v 0 = v* + 0.7 V. 
For Vf = -20 V, v 0 = -20 V + 0.7 V = -19.3 V 
For Vi = -0.7 V, v Q = -0.7 V + 0.7 V = 0 V 




17 



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(b) For Vi < 5 V the 5 V battery will ensure the diode is forward-biased and v 0 = v,- - 5 V. 
At v/ = 5 V 

v o = 5V-5V = 0V 
At V/ = -20 V 
v 0 = -20 V - 5 V = -25 V 

For Vi > 5 V the diode is reverse-biased and v 0 = 0 V. 




33. (a) Positive pulse of vf. 

i.2knqov-o.7V) 

1.2 kQ + 2.2 kQ 
Negative pulse of v{. 
diode “open”, v 0 = 0 V 

(b) Positive pulse of vf. 

Vo = 10 V - 0.7 V + 5 V = 14.3 V 
Negative pulse of v{. 
diode “open”, v Q = 0 V 



34. (a) For v* = 20 V the diode is reverse-biased and v 0 = 0 V. 

For Vi = -5 V, Vi overpowers the 2 V battery and the diode is “on”. 

Applying Kirchhoff s voltage law in the clockwise direction: 

-5V + 2V-v o = 0 

v G = -3 V 



(b) For Vi = 20 V the 20 V level overpowers the 5 V supply and the diode is “on”. Using the 
short-circuit equivalent for the diode we find v a = v t = 20 V. 

For Vi = -5 V, both v* and the 5 V supply reverse-bias the diode and separate v t from v Q . 
However, v 0 is connected directly through the 2.2 kQ resistor to the 5 V supply and 
v 0 = 5 V. 






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35. (a) Diode “on” for v* > 4.7 V 

For v f > 4.7 V, V Q = 4 V + 0.7 V = 4.7 V 
For Vf < 4.7 V, diode “off’ and v 0 = v* 

(b) Again, diode “on” for v* > 4.7 V but v 0 

now defined as the voltage across the diode 
For v f > 4.7 V, v 0 = 0.7 V 




For Vf < 4.7 V, diode “off’, I D = I R = 0 mA and V 22kn = IR = ( 0 mA)R = 0 V 



Therefore, v 0 = V/-4V 
At Vi = 0 V, v 0 = -4 V 
Vi = - 8 V, v 0 = -8 V - 4 V = -12 V 




36. For the positive region of v t : 

The right Si diode is reverse-biased. 

The left Si diode is “on” for levels of v t greater than 
5.3 V + 0.7 V = 6 V. In fact, v 0 = 6 V for v, > 6 V. 

For Vi < 6 V both diodes are reverse-biased and v Q = v*. 

For the negative region of v,-: 

The left Si diode is reverse-biased. 

The right Si diode is “on” for levels of v z more negative than 7.3 V + 0.7 V = 8 V. In 
fact, v 0 = -8 V for v* < -8 V. 



For Vi > -8 V both diodes are reverse-biased and v 0 = v t . 




i R : For -8 V < v t < 6 V there is no conduction through the 10 kQ resistor due to the lack of a 
complete circuit. Therefore, i R = 0 mA. 

For Vi > 6 V 

VR = Vi - v 0 = Vi - 6 V 

For v/ = 10 V, v* = 10 V - 6 V = 4 V 
4 V 

and i R = = 0.4 mA 

10 kQ 

For Vi < -8 V 

V R = Vi - v 0 = Vi + 8 V 



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For Vi = -10 V 

v R = -10 V + 8 V = -2 V 
-2 V 

and i R = = -0.2 mA 

10 kQ 

*R a 



+6 V 


V 

- 

• 0.4'mA v 


T 


\ v -04mA 








-8V \* ‘ 



37. (a) Starting with v* = -20 V, the diode is in the “on” state and the capacitor quickly charges 

to -20 V+. During this interval of time v 0 is across the “on” diode (short-current 
equivalent) and v 0 = 0 V. 

When Vi switches to the +20 V level the diode enters the “off’ state (open-circuit 
equivalent) and v 0 = v* + v c = 20 V + 20 V = +40 V 



total swing of = total swing of v/ 

(b) Starting with v,- = -20 V, the diode is in the “on” state and the capacitor quickly charges 
up to -15 V+. Note that v t = +20 V and the 5 V supply are additive across the capacitor. 
During this time interval v 0 is across “on” diode and 5 V supply and v 0 = -5 V. 

When Vi switches to the +20 V level the diode enters the “off’ state and v 0 = v,- + v c = 

20 V+ 15 V = 35 V. 



40 V swing (= that of u j ) 





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38. (a) For negative half cycle capacitor charges to peak value ofl20V-0.7V=119.3V with 

polarity (— — 1( 1-). The output v a is directly across the “on” diode resulting in 

v 0 = -0.7 V as a negative peak value. 

For next positive half cycle v 0 = v,- + 1 19.3 V with peak value of 
v 0 = 120 V + 119.3 V = 239.3 V. 



vertical shift of 119.3 V 



(b) For positive half cycle capacitor charges to peak value of 120 V - 20 V - 0.7 V = 99.3 V 
with polarity (H — — 1( — — ) . The output v 0 = 20 V + 0.7 V = 20.7 V 
For next negative half cycle v 0 = v,- - 99.3 V with negative peak value of 
v 0 = -120 V - 99.3 V = -219.3 V. 

| vertical shift of -99.3 V 

99.3 V 



Using the ideal diode approximation the vertical shift of part (a) would be 120 V rather 
than 1 19.3 V and -100 V rather than -99.3 V for part (b). Using the ideal diode 
approximation would certainly be appropriate in this case. 

39. (a) r= 7?C= (56 kQ)(0.1 //F) = 5.6ms 

5 t = 28 ms 

,1 x . r lms 

(b) 5t = 28 ms » — = = 0.5 ms, 56:1 

2 2 

(c) Positive pulse of v*: 

Diode “on” and v 0 = -2 V + 0.7 V = -1.3 V 
Capacitor charges tol0V + 2V-0.7V=11.3V 

Negative pulse of v,-: 

Diode “off’ and v 0 = -10 V - 1 1.3 V = -21.3 V 



-1.3 






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40. 



Solution is network of Fig. 2.176(b) using a 10 V supply in place of the 5 V source. 



41. Network of Fig. 2.178 with 2 V battery reversed. 




42. (a) In the absence of the Zener diode 

1 80 0(20 V) =9y 
180Q + 220Q 

Vl = 9V < Vz= 10V and diode non-conducting 
20 V 

Therefore, I L = I R = = 50 mA 

220 Q + 180Q 

with // = 0 mA 
and V L = 9 V 

(b) In the absence of the Zener diode 

470 Q(20 V) = 13 62y 
470 Q + 220 Q 

V L = 13.62 V > V z = 10 V and Zener diode “on” 



Therefore, V L = 10 V and F* = 10 V 

I Rs = V R s /R s = 10 V/220 Q = 45.45 mA 

h = V l /R l = 10 V/470 Q = 21.28 mA 
and I z = I Rs - I L = 45.45 mA- 21.28 mA = 24.17 mA 



(c) P Zimx = 400 mW =V Z I Z =( 10 V)(/ z ) 

r 400 mW .. . 

I z = = 40 mA 



= I* 



10 V 
Ir s ~ 

= 10 V = 1,834.86 Q 



I z = 45.45 m A - 40 mA = 5.45 mA 



5.45 mA 



Large R L reduces I L and forces more of I R to pass through Zener diode. 



(d) 



In the absence of the Zener diode 

tW0V= ^ (20V) 
R l + 220 Q 

10 R l + 2200 = 20 R l 
10 R l = 2200 

R l = 220 Q 



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12 V 



= 60 Q 



43. 



(a) V z = 12V,R l =^ = 
V L =V Z = 12V = 



I L 200 mA 



7^ _ 60 Q(16 V) 



7?^ + 7? o 60 £2 + 7? ( 

720 + 127?, = 960 
127?, = 240 
7? v = 20 Q 



44. 



45. 



(b) P z _ = V Z I Z _ 

= (12 V)(200 mA) 

= 2.4 W 

V V 

Since I L = — T - = — is fixed in magnitude the maximum value of I R will occur when I z is a 
R l R l 

maximum. The maximum level of I R will in turn determine the maximum permissible level 

off,. 



U =- 



Il = 



400 mW 



V z 8 V 
V r V 7 8 V 



= 50 mA 



= 36.36 mA 



R l R l 220 Q 

I R = 7z + A = 50 mA + 36.36 mA = 86.36 mA 

V — V 
I = Vi Vz 

Rs R s 

or Vi= I r R s + F z 

= (86.36 mA)(91 Q) + 8 V = 7.86 V + 8 V = 15.86 V 

Any value of v* that exceeds 15.86 V will result in a current I z that will exceed the maximum 
value. 

At 30 V we have to be sure Zener diode is “on”. 



.*. V L = 20 V = 



_ R L V i _ 1 kQ(30 V) 



7? j + 7? , 1 kf2 + 7? 

L s s 



Solving, 7?, = 0.5 kQ 



, r 50V -20V ^ _ 20V ^ . 

At 50 V, I R = = 60 mA, I L = = 20 mA 

0.5 kQ 1 kQ 



Izm = I R ~ I i. = 60 mA - 20 mA = 40 mA 



46. For V/ = +50 V: 

Z\ forward-biased at 0.7 V 

Z 2 reverse-biased at the Zener potential and V z ^ =10 V. 
Therefore, V Q = K Zi + K Zz = 0.7 V + 10 V = 10.7 V 



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For Vi = -50 V: 

Z\ reverse-biased at the Zener potential and F Z| = -10 V. 

Z 2 forward-biased at -0.7 V. 

Therefore, V Q = K Z| + F z? = -10.7 V 



4^o 



z 



IO.-7V 



Z 



-tonv 



For a 5 V square wave neither Zener diode will reach its Zener potential. In fact, for either 
polarity of v t one Zener diode will be in an open-circuit state resulting in v Q = v t . 



AT. ' 


' 5V 






Z 




~o 




z 








-sv 



47. V m = 1.414(120 V) = 169.68 V 

2V m = 2(169.68 V) = 339.36 V 

48. The PIV for each diode is 2V m 

•'•PIV = 2(1.414)(F rms ) 



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Chapter 3 



1. 

2. A bipolar transistor utilizes holes and electrons in the injection or charge flow process, while 
unipolar devices utilize either electrons or holes, but not both, in the charge flow process. 

3. Forward- and reverse-biased. 

4. The leakage current I C o is the minority carrier current in the collector. 

5. 

6 . 



7. 



8. A the largest 
I B the smallest 
Ic = h 

9. h 
Ie 
Ib 
Ic 

10 . 

11. 4= 5 mA, Vcb= IV: V BE = 800 mV 

V CB = 10 V: V BE = 770 mV 
V CB = 20 V: V BE = 750 mV 

The change in V CB is 20 V: 1 V = 20:1 

The resulting change in V B e is 800 mV: 750 mV =1.07:1 (very slight) 



100 



I C ^>I C = 100I B 



= Ic + Ib=100Ib + Ib=10Ub 

/,, 8 mA __ 4 

= — ^ = = 79.21 uA 

101 101 

= 100/j = 100(79.21 juA) = 7.921 mA 



12 . 



(a) 

(b) 



AV 
A I 



0.9 V- 0.7 V 



= 25 Q 



8 mA-0 

Yes, since 25 Q is often negligible compared to the other resistance levels of the 
network. 



13. (a) I c = I E = 4.5 mA 

(b) I C = I E = 4.5mA 

(c) negligible: change cannot be detected on this set of characteristics. 

(d) I c = Ie 



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14. (a) Using Fig. 3.7 first, I E = 7 mA 

Then Fig. 3.8 results in I c = 7 mA 

(b) Using Fig. 3.8 first, I E = 5 mA 
Then Fig. 3.7 results in F^ = 0.78 V 

(c) Using Fig. 3.10(b) I E = 5 mA results in F^ = 0.81 V 

(d) Using Fig. 3.10(c) I E = 5 mA results in V BE = 0.7 V 

(e) Yes, the difference in levels of V BE can be ignored for most applications if voltages of 
several volts are present in the network. 

15. (a) I c = a I E = (0.998)(4 mA) = 3.992 mA 



(b) I E = I c + I B => Ic — Ie~ Ib = 2.8 mA - 0.02 mA = 2.78 mA 
2.78 mA 



dc I E 2.8 mA 



= 0.993 



(c) I C = PI B = 



a 



lc 

a 



1 - a 
1.96 mA 
0.993 



I B = 



0.98 

1-0.98 



(40 juA) = 1.96 mA 



16. 

17. 



It = VJRt = 500 mV/20 Q = 25 mA 
I L = Ii = 25 mA 

V L = IlRl = (25 mA)(l kQ) = 25 V 
V 25 V 

A v = ^ = =50 

V, 0.5 V 



V. 



200 mV 



200 mV 



18. I f = 

Ri+R s 20Q + 100Q 120 Q 

I L = Ii= 1.67 mA 

V l = IlR = (1.67 mA)(5 kQ) = 8.35 V 

V 8 35 V 

^ = =41.75 

V, 0.2 V 



= 1.67 mA 



19. 










20. (a) 


Fig. 


3.14(b): 


Ib = 


35/uA 




Fig. 


3.14(a): 


Ic = 


3.6 mA 


(b) 


Fig. 


3.14(a): 


V C E 


= 2.5 V 




Fig. 


3.14(b): 


Vbe 


= 0.72 V 



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21 . 



(a) 

(b) a = 



7, = 2mA =117 « 5 
I R 17 juA 



L B 

P 



117.65 



= 0.992 



P + 1 117.65 + 1 

(c) I ceo = 0.3 mA 

(A) IcBO = (1 — ofylcEO 

= (1 - 0.992)(0.3 mA) = 2.4 //A 



22. (a) Fig. 3.14(a): I ceo = 0.3 mA 

(b) Fig. 3.14(a): I c = 1.35 mA 
135mA 
I B 10 //A 



(c) a= -?— = —= 0.9926 
J3 + 1 136 

IcBO = (1 “ OC)IcEO 

= (1 - 0.9926)(0.3 mA) 

= 2.2 juA 



23. 



, , I ( 6.7 mA 

(a) /? dc = = = 83.75 

/ s 80 juA 

o E 0.85 mA 

(b) ldc=+= _ . =170 

5 //A 

, v n _ 1 C 3 ’ 4mA 

( c ) Ac _ A 113.33 

I B 30 //A 

(d) /? dc does change from pt. to pt. on the characteristics. 
Low I B , high V C e —> higher betas 

High Ib ? low Fc# — > lower betas 



24. 



(a) 

M b 

(b) Ac= 77 - 

m b 



(c) 



7.3 mA - 6 mA 1.3mA 






V CE = 15 V 



= 65 



Ac A/„ 



90 /uA - 70 //A 20 juA 

1.4 mA- 0.3 mA _ 1.1mA 
10 /uA - 0 /uA 10 juA 

4.25 mA- 2.35 mA 1.9 mA 



Vce = 10 V 



110 



= 95 



40 /uA - 20 /uA 20 juA 

(d) /? ac does change from point to point on the characteristics. The highest value was 
obtained at a higher level of V C e and lower level of I c . The separation between I B 
is the greatest in this region. 



27 

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curves 




VcE 


h 


Pdc 


P * c 


Ic 


PdJ Pac 


5 V 


80 juA 


83.75 


65 


6.7 mA 


1.29 


10 V 


30 /uA 


113.33 


95 


3.4 mA 


1.19 


15 V 


5 /uA 


170 


110 


0.85 mA 


1.55 



As I c decreased, the level of /? dc and j3 ac increased. Note that the level of /? dc and p ac in 
the center of the active region is close to the average value of the levels obtained. In 
each case /? dc is larger than p ac , with the least difference occurring in the center of the 
active region. 



25. 



= I c 2.9 mA 
Pdc I B 25 //A 

= P - 116 

^ p + 1 _ 116 + 1 



= 116 



= 0.991 



I E = Ida= 2.9 mA/0.991 =2.93 mA 



26. 



(a) p= 



a 



0.987 0.987 



(b) a = ^ 



l-a 1-0.987 0.013 

120 120 



= 75.92 



>0 + 1 120 + 1 121 



= 0.992 



( C ) i B =I<L=?^=n.njuA 

pm H 

Ie = I C + Ib = 2mA+ 11.11 juA 

= 2.011mA 



27. 

28. 



V e = V i -V be = 2Y-0AY= 1.9 V 
V 1.9 V 

^ v = -^ = ^22 =0.95= 1 
V t 2 V 

r V E 1-9 V , 

4 = — = = 1.9 mA (rms) 

74 1 kQ 



29. 



Output characteristics: 



c CE 



B 




= > 







■> 



Curves are essentially the 
same with new scales as shown. 



Input characteristics: 

Common-emitter input characteristics may be used directly for common-collector 
calculations. 



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30. 



P r = 30 mW = V CE Ic 



Ic=I c >Vce = 



V CE =V CEmax ,Ic = 
V CE = 10 V,/c = 
I c = 4 rnA, V CE = 
Vce=15V,I c = 



p c„ 


_ 30 mW 


7 C,„ 


7mA 


. 7 C„ 


_ 30 nA\ 




20 V 


p c,» 


_ 30 mW _ 


V CE 


10 V 


Pc m 


_ 30 mW 


Ic 


4 mA 


p c„„ 


30 mW _ 




15 V 



= 4.29 V 



= 1.5 mA 



= 3 mA 



= 7.5 V 



= 2 mA 




31. /c = I Cmax , V CE = 

V C B=V CBmax ,Ic = 
Ic = 4 mA, Fes = 
Fes = 10 V, I c = 



p c n ,» 


_ 30 mW 


Ic^ 


6 mA 


. p c, . 


30 mW 


V CB m 


15 V 


P Cm 


_ 30 mW _ 


Ic 


4 mA 


p c„„ 


30 mW _ 


PCS 


10 V 



= 5 V 



= 2 mA 



= 7.5 V 
= 3 mA 




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32. 



The operating temperature range is -55°C < Tj< 150°C 

°F= -°C +32° 

5 

= — (-55°C) + 32° = -67°F 
5 

°F = j(150°C) + 32° = 302°F 
. . -67°F< r / <302°F 



33. 



I c = 200 mA, V CE = 30 V, P D = 625 mW 



Ic= I Cm ,V CE = 
v ce= V CE _,I C = 

Ic = 100 mA, V ce = 
V CE = 20 V,/ c = 





_ 625 mW 




200 mA 


. P » m 


_ 625 mW 


V CE m 


30 V 


_P Dm 


625 mW 


* Ic 


“ 100 mA 


P D m _ 


625 mW _ 


VCE 


20 V 



= 3.125 V 



= 20.83 mA 



= 6.25 V 



= 31.25 mA 




34. 



From Fig. 3.23 (a) I C bo = 50 nA max 



P*s 



Pm in + Pm 

2 

50 + 150 



200 

2 



= 100 



:.Iceo = PIcbo = ( 100)(50nA) 

= 5 juA 



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35 . 



h FE (Ac) with V CE = 1 V, T 7 = 25°C 
/c = 0.1 mA, 0.43(100) = 43 

i 

I c = 10 mA, h FE = 0.98(100) = 98 



h fe (J3 a c ) with = 10 V, T= 25°C 

/( = 0. 1 mA, h fe = 72 

Ic = 10 mA, hfe =160 

For both h FE and /z/ e the same increase in collector current resulted in a similar increase 
(relatively speaking) in the gain parameter. The levels are higher for h fe but note that V CE is 
higher also. 



36. 



As the reverse-bias potential increases in magnitude the input capacitance C ibo decreases (Fig. 
3.23(b)). Increasing reverse-bias potentials causes the width of the depletion region to 



increase, thereby reducing the capacitance 




37. (a) At A = 1 mA, hf e = 120 

At 7c = 10 mA, hf e = 160 

(b) The results confirm the conclusions of problems 23 and 24 that beta tends to increase 
with increasing collector current. 



39. 



(a) Ac= YT 

A /„ 



1 6 mA - 1 2.2 mA 3.8 mA 



Vce= 3 V 



80 /uA - 60 /jA 20 juA 



= 190 



(b) Ac 

(c) Ac 

(d) Ac 



I c 12 mA 
I B 59.5 juA 

4 mA - 2 mA 
18 juA -8 /uA 

I c _ 3 mA 
I B 13 juA 



= 201.7 

2 mA 
~ 10 /uA 

230.77 



= 200 



(e) In both cases Ac is slightly higher than Ac (= 10%) 



(f>(g) 

In general Ac and Ac increase with increasing I c for fixed V CF and both decrease for 
decreasing levels of V CF for a fixed I E . However, if I c increases while V CE decreases 
when moving between two points on the characteristics, chances are the level of Ac or 
Ac may not change significantly. In other words, the expected increase due to an 
increase in collector current may be offset by a decrease in V CE . The above data reveals 
that this is a strong possibility since the levels of / 3 are relatively close. 



31 

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Chapter 4 



1. 


(a) 


T _Vcc~ 

Bq r, 




(b) 


V, = P\ 




(c) 


o 

ii 

r Ol 




(d) 


II 

to 




(e) 


II 

£ 

II 

£ 




(f) 


V E = 0 V 


2. 


(a) 


Ic = Ph = 8 




(b) 


V R 

Rc=-^ = 

* c 



16 V-0.7 V 15.3 V 



470 kQ 



470 kQ 



= 32.55 juA 



6 V 



3.2 mA 3.2 mA 



= 1.875 kQ 



(c) R b =^ = 



12 V-0.7 V 11.3V 



40 /uA 40 juA 



= 282.5 k Q 



(d) V CE =Vc = 6 V 

(a) Ic = Ie~ Ib = 4 mA - 20 //A = 3.98 mA = 4 mA 

(b) V cc = V C e + IcRc = 7.2 V + (3.98 mA)(2.2 kQ) 

= 15.96 V= 16 V 



(c) p= 



3.98 mA 
20 juA 



= 199 = 200 



(d) R b = 



V r r V cc -V Bl 



h h 



15.96 V -0.7 V 

20 juA 



= 763 kQ 



I, = 



v 

_ y cc 



16 V 



R c 2.7 kQ 



= 5.93 mA 



(a) Load line intersects vertical axis at I c = 
and horizontal axis at Vce = 2 1 V. 



21 V 
3 kQ 



= 7 mA 



v - V 21V-07V 

(b) I B = 25 juA: R b = V cc = — U 7 V = 812 kQ 
I B 25 juA 



(c) 7 C = 3.4 mA, = 10.75 V 



32 

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(d) P= 'f 

*B 



3.4 mA 
25 juA 



136 



(e) a = 



(f) 



p _ 136 _ 136 
P + \~ 136 + 1 _ 137 

21V 



= 0.992 



v 

T = cc 

1 n 



R r 3 kQ 



= 7 mA 



(g) - 

(h) P D = V CEq I Cq = (10.75 V)(3.4 mA) = 36.55 mW 



(i) P s = Vcdlc + I B ) = 21 V(3.4 mA + 25 juA) = 71.92 mW 

(j) Pr = P s ~Pd = 71.92 mW - 36.55 mW = 35.37 mW 



(a) / v cc-Vbe _ 20 V-0.7 V _ 19.3 V 

K Bq R B +(p + \)R E 510 kQ + (101)1.5 kQ 661.5 kQ 

= 29.18 juA 

(b) I Cq = PI Bq = (100)(29.18 //A) = 2.92 mA 



( c ) v ce q = Vcc - IdRc + R e ) = 20 V - (2.92 mA)(2.4 kQ + 1 .5 kQ) 

= 20 V- 11.388 V 

= 8.61 V 

(d) F c = F cc - /c^c = 20 V - (2.92 mA)(2.4 kQ) = 20 V - 7.008 V 

= 13 V 

(e) V B = Fee - hRs = 20 V - (29. 1 8 //A)(5 1 0 kQ) 

= 20 V- 14.882 V = 5.12 V 

(f) Fe= F c - F c ^= 13 V - 8.61 V = 4.39 V 



7. 



(a) 




12 V-7.6 V 
2 m A 



4.4 V 
2 mA 



= 2.2 kQ 



F 2 4V 

(b) 4-/ c : Re=-^ = —— =1.2kQ 
A 2 mA 



(C) 




v -V -V 

y CC y BE y E 



12 V-0.7 V-2.4V 
2 mA/80 



8.9 V 
25 //A 



= 356 kQ 



(d) F C £= F c - V e = 7.6 V - 2.4 V = 5.2 V 

(e) V B =V BE + V E = 0.7 V + 2.4 V = 3.1 V 



33 

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2.1 V 



= 3.09 mA 



(a) 



Ic,I E ^ 



R e 0.68 kQ 
= 109mA , 
I B 20 juA 



(b) V cc =V Rc +Vce+Ve 

= (3.09 mA)(2.7 kQ) + 7.3 V + 2.1 V = 8.34 V + 7.3 V + 2.1 V 

= 17.74 V 



(c) r b= V *s _ Vcc-Vbe-V e _ 17.74 V-0.7V-2.1V 



± B 

14.94 V 

20 juA 



20 //A 



= 747 kQ 



Ir =~ 



20 V 



20 V 



i? c +7^ 2.4 kQ + 1.5 kQ 3.9 kQ 



= 5.13 mA 



10 . 



(a) I c = 6.8 mA = 



24 V 



R E + R E R e +1.2 kQ 



24 V 

7? c + 1.2 kQ= — = 3.529 kQ 

6.8 mA 

R c = 2.33 kQ 



(b) P~ — = -r^-— — — - ,33.33 



I B 30 /jA 



(C) 



Rb = 



v,_ 



V -V -V 

, V CC y BE V E 



18.5 V 



30 /uA 



= 616.67 kQ 



24 V - 0.7 V - (4 mA)(l .2 kQ) 
30 /uA 



(d) Pd V CEq I Cq 

= (10 V)(4 mA) = 40 mW 

(e) P = I 2 c R c = (4 mA) 2 (2.33 kQ) 

= 37.28 mW 



11. (a) Problem 1: I Cq = 2.93 mA, V CEq = 8.09 V 

(b) I Bq = 32.55 /uA (the same) 

I Cq = pI Bo = (135)(32.55 /uA) = 4.39 mA 

V CEq = V cc ~I Co R c = 16 V - (4.39 mA)(2.7 kQ) = 4.15 V 



34 

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(c) %Mc = 



4.39 mA -2.93 mA 



%A V CE = 



2.93 mA 
4.15 V-8.09 V 



x 100% = 49.83% 



x 100% = 48.70% 



8.09 V 

Less than 50% due to level of accuracy carried through calculations, 
(d) Problem 6: I c = 2.92 mA, V CEq = 8.61 V (/ = 29.18 juA) 

20V-0.7 V 



V -V 

(e) I B = ^ 

q R B +{p+m 



510kQ + (150 + l)(1.5kQ) 

I c = pI Bo = (150)(26.21 juA) = 3.93 mA 

V CEq =V cc -Ic(Rc + Re) 

= 20 V - (3.93 mA)(2.4 kQ + 1.5 kQ) = 4.67 V 
3.93 mA -2.92 mA 



= 26.21 juA 



(f) %A I c = 
%A V CE = 



2.92 mA 
4.67 V-8.61 V 



x 100% = 34.59% 



x 100% = 46.76% 



8.61 V 

(g) For both I c and V C e the % change is less for the emitter-stabilized. 



? 

12. pR E > 10R 2 

(80)(0.68 kQ)> 10(9.1 kQ) 

54.4 kQ ^ 91 kQ (No!) 

(a) Use exact approach: 

R m = Ri || Ri = 62 kQ || 9.1 kQ = 7.94 kQ 
^ L= (9.1kQ)(16V) =205V 
R 2 + R } 9.1 kQ + 62 kQ 
J Eth-Vbe _ 2,05 V-0.7 V 

R Th + (P + 1)^ 7.94 kQ + (8 1)(0.68 kQ) 

= 21.42 juA 

(b) I Cq = PI Bq = (80X21.42 JUA) = 1.71 mA 

(c) V CEq =V cc ~ I Cq (Rc + Re) 

= 16 V- (1.71 mA)(3.9 kQ + 0.68 kQ) 

= 8.17 V 

(d) V c =Vcc-IcRc 

= 16 V- (1.71 mA)(3.9 kQ) 

= 9.33 V 

(e) V E = IeRe = IcRe = (L71 mA)(0.68 kQ) 

= 1.16 V 

(f) V B = V E + V BE = 1.16 V + 0.7 V 

= 1.86 V 



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13. 



(a) 




18 V-12V 
4.7 kQ 



= 1.28 mA 



(b) V E = I e Re = IcRe = (1 .28 mA)(l .2 kQ) = 1.54 V 

(c) Vb — V BE + Ve = 0.7 V + 1.54 V = 2.24 V 



(d) Ri = -^: V Ri = V cc -Vb=1SV- 2.24 V = 15.76 V 

T T V B 2.24 V _ „ . 

/,, = I p = — = = 0.4 mA 



Ri = 



15.76 V 
0.4 mA 



R 2 5.6 kQ 



= 39.4 kQ 



14. (a) I c = ph = (100)(20 juA) = 2 mA 

(b) I E = I c + I b = 2 mA + 20 juA 

= 2.02 mA 

V E = I e Re = (2.02 mA)(l .2 kQ) 

= 2.42 V 

(c) V cc =V C + I C R C = 10.6 V + (2 mA)(2.7 kQ) 

= 10.6 V+ 5.4 V 

= 16 V 



(d) V CE = V c - V E = 10.6 V- 2.42 V 

= 8.18 V 

(e) V B =V E + V B e = 2.42 V + 0.7 V = 3.12 V 



(f) I R l = I r 2 + h 

= 3.12 V 

8.2 kQ 
v - V 

= CC y B 



+ 20 /uA = 380.5 /uA + 20 juA = 400.5 juA 



16 V-3.12 V 
400.5 //A 



= 32.16 kQ 



15. 



Ec 16 V 

+ R e 3.9 kQ + 0.68 kQ 



16 V 
4.58 kQ 



= 3.49 mA 



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16. (a) pR E >lOR 2 

(120)(1 kQ) > 10(8.2 kQ) 
120 kQ > 82 kQ (checks) 



r,v„ 



(8.2 kQ)(18 V) 



= 3.13 V 



R y + 7?2 39 kQ + 8.2 kQ 

V E = V b - V be = 3.13 V- 0.7 V = 2.43 V 



I c = Ie= e 



2.43 V 



R e 1 kQ 



= 2.43 mA 



(b) V ce Vcc ~ Ic(Rc + Re) 

= 18 V - (2.43 mA)(3.3 kQ + 1 kQ) 

= 7.55 V 



17. 



18. 



I c 2.43 mA 

(c) I B = — = = 20.25 juA 

P 120 ^ 



(d) V E = I e Re = IcRe = (2.43 mA)(l kQ) = 2.43 V 

(e) V B = 3.13 V 

(a) R rh = Ri || R 2 = 39 kQ || 8.2 kQ = 6.78 kQ 
RrVrr 8.2kQ(18V) 



- Th - 



Ib = 



= 3.13 V 

R y + 7?2 29 kQ + 8.2 kQ 

E Th -V BE _ 3.13 V-0.7 V 



R rh +(p + 1 )R e 6.78 kQ + (12 1)(1 kQ) 

2 43 V 

= — : = 19.02 juA 

127.78 kQ 

Ic = Ph = (120)(19.02 juA) = 2.28 mA (vs. 2.43 mA #16) 

(b) V CE = V cc ~ IdRc + R e ) = 1 8 V - (2.28 mA)(3.3 kQ + 1 kQ) 

= 18 V - 9.8 V = 8.2 V (vs. 7.55 V #16) 

(c) 19.02 juA (vs. 20.25 juA #16) 

(d) V E = I e Re = IcRe = (2.28 mA)(l kQ) = 2.28 V (vs. 2.43 V #16) 

(e) V b =V be + V e = 0.7 V + 2.28 V = 2.98 V (vs. 3.13 V #16) 

The results suggest that the approximate approach is valid if Eq. 4.33 is satisfied. 

/xr, ^2 t/~ 9.1 kQ(16 V) , 

(a) V B = — V rr = = 2.05 V 

^+^ 2 62 kQ + 9.1 kQ 

V E = V b - V be = 2.05 V-0.7 V= 1.35 V 
1.35 V 
R e 0.68 kQ 
I r =I E = 1.99 mA 



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(b) 



V CEq =V cc -Ic{Rc + Re) 

= 16 V - (1.99 mA)(3.9 kQ + 0.68 kQ) 
= 16 V-9.11 V 

= 6.89 V 



I Cq _ 1.99 mA 

~/T _ 80 



24.88 juA 



From Problem 12: 

I r = 1.71 mA, V CF = 8.17 V, I R = 21.42 juA 

Cg 7 c tLg 7 u q 



(c) The differences of about 14% suggest that the exact approach should be employed when 
appropriate. 



19. 



(a) I c = 7.5 mA = 



V r < 



r c +r e 



24 V 
3 R e + Re 



24 V 
4R, 



Re 



24 V 



24 V 



= 0.8 kQ 



4(7.5 mA) 30 mA 
R c = 3 R e = 3(0.8 kQ) = 2.4 kQ 

(b) V E = I e Re = IcRe = (5 mA)(0.8 kQ) = 4 V 

(c) V B = V E + V be = 4 V + 0.7 V = 4.7 V 

R V 

(d) v B = K2Vcc 



R 2 + R] 



4.7 V = 



_ (24 V) 



R 2 + 24 kQ 



R 2 = 5.84 kQ 



(e) /?dc= —= 5mA = 129.8 
I B 38.5 juA 



(f) fiR E >\0R 2 

(129.8)(0.8 kQ) > 10(5.84 kQ) 

103.84 kQ > 58.4 kQ (checks) 

20. (a) From problem 12b, I c = 1.71 mA 

From problem 12c, V C e = 8.17 V 

(b) p changed to 120: 

From problem 12a, E Th = 2.05 V, R rh = 7.94 kQ 
j _ E Th -V BE 2,05 V-0.7 V 

5 R Th +(p + \)R E 7.94 kQ + (121)(0.68 kQ) 

= 14.96 /uA 

Ic = Ph = (120)(14.96 juA) = 1.8 mA 
Vqe = Vcc ~ Ic(Rc + Re) 

= 16 V - (1.8 mA)(3.9 kQ + 0.68 kQ) 

= 7.76 V 



38 



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(C) 



X 100% = 5.26% 



%AI C = 
%AV ce : 



1.8 mA-1.71 mA 
1.71 mA 

_ 7.76 V-8.17 V 
8.17 V 



x 100% = 5.02% 



(d) 



%AI C 
%A V CE 



11c 

49.83% 

48.70% 



Fixed-bias 



Ilf 

34.59% 

46.76% 



Emitter 

feedback 



20c 

5.26% 

5.02% 



Voltage- 

divider 



(e) Quite obviously, the voltage-divider configuration is the least sensitive to changes in p. 

21. I. (a) Problem 16: Approximation approach: I c = 2.43 mA, V CE = 7.55 V 

Problem 17: Exact analysis: I c = 2.28 mA, V CE = 8.2 V 

The exact solution will be employed to demonstrate the effect of the change of p. Using 
the approximate approach would result in %AI C = 0% and %A V C e = 0%. 



(b) Problem 17: E Th = 3.13 V, R Th = 6.78 kQ 

E th -V be _ 3.13 V -0.7 V _ 2.43 V 

5 R n +{p + 1 )R e ~ 6.78 kQ + (180 + 1)1 kQ _ 187.78 kQ 
= 12.94 juA 

Ic = Ph = (180)( 12.94 juA) = 2.33 mA 

Vce = V cc ~ Ic(Rc + Re) = 18 V- (2.33 mA)(3.3 kQ + 1 kQ) 

= 7.98 V 



(c) %A I c = 
%AV ce = 



2.33 mA - 2.28 mA 



2.28 mA 
7.98 V- 8.2 V 
8.2 V 



x 100% = 2.19% 



x 100% = 2.68% 



For situations where pR E > 107? 2 the change in I c and/or V C e due to significant change in 
p will be relatively small. 

(d) %A I c = 2.19% vs. 49.83% for problem 1 1 . 

%A V C e = 2.68% vs. 48.70% for problem 11. 

(e) Voltage-divider configuration considerably less sensitive. 

II. The resulting %AI C and %AV C e will be quite small. 



39 

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16 V-0.7V 



R b + P(R C +R e ) 470 kQ + (120)(3.6 kQ + 0.51 kQ) 

= 15.88 juA 

(b) I c = ph = (120)(15.88 juA) 

= 1.91 mA 

(c) Vc = Vcc ~ IcRc 

= 16 V- (1.91 mA)(3.6 kQ) 

= 9.12 V 



23. 



, , , V cc - V BE 30V-0.7V 

(a) 1 B = — — = 

R b + p(R c + R e ) 6.90 kQ + 100(6.2 kQ + 1.5 kQ) 



20.07 juA 



I c = Ph = (100)(20.07 juA) = 2.01 mA 

(b) V c =Vcc-IcRc 

= 30 V - (2.01 mA)(6.2 kQ) = 30 V - 12.462 V = 17.54 V 

(c) V E = I e Re = IcRe = (2.01 mA)(l .5 kQ) = 3.02 V 



(d) V CE = V cc ~ Ic(Rc + R e ) = 30 V - (2.0 1 mA)(6.2 kQ + 1 .5 kQ) 

= 14.52 V 



24. 



(a) 



(b) 



Vcc-Vre _ 22 V-0.7 V 

R b + P(R C + R e ) 470 kQ + (90)(9. 1 kQ + 9. 1 kQ) 



= 10.09 juA 

Ic = Ph = (90)( 10.09 juA) = 0.91 mA 

V CE = V cc ~ Ic(Rc + R E ) = 22 V- (0.9 1 mA)(9. 1 kQ + 9. 1 kQ) 

= 5.44 V 



P= 135, I B = 



V -V 

V CC V BE 



22 V-0.7 V 



R b + P(R C + R e ) 470 kQ + (135)(9. 1 kQ + 9. 1 kQ) 



= 7.28 juA 



Ic = Ph = (135)(7.28 juA) = 0.983 mA 

Vce = Vcc - Ic(Rc + R E ) = 22 V- (0.983 mA)(9. 1 kQ + 9. 1 kQ) 

= 4.11 V 



(c) %A I c = 



0.983 mA -0.9 1mA 



%AV ce = 



0.91 mA 
4.11 V-5.44 V 



x 100 % = 8 . 02 % 



5.44 V 



x 100% = 24.45% 



(d) The results for the collector feedback configuration are closer to the voltage-divider 
configuration than to the other two. However, the voltage-divider configuration 
continues to have the least sensitivities to change in p. 



40 



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25. 1 MQ = 0 Cl, R b = 150 kQ 

j _ Vcc-Vee _ 12 V-0.7 V 

^ R b + J3(R C +R e ) 150 kQ + (180)(4.7 kQ + 3.3 kQ) 
= 7.11 juA 

Ic = PIb = ( 180)(7.11 juA) = 1.28 mA 
V c = V cc -IcRc = 12 V - (1 .28 mA)(4.7 kQ) 

= 5.98 V 

Full 1 MQ: R b = 1,000 kQ + 150 kQ = 1,150 kQ = 1.15 MQ 

J _ Vcc-Vre 12 V-0.7 V 

B R b + J3(R C + R e ) 1 . 1 5 MQ + (1 80)(4.7 kQ + 3 .3 kQ) 

= 4.36 juA 

I c = pI B = (180)(4.36 juA) = 0.785 mA 

V c = V cc ~ IcRc = 12 V - (0.785 mA)(4.7 kQ) 

= 8.31 V 

V c ranges from 5.98 V to 8.31 V 



26. 


(a) 


V E = V b -V be = 4 V-0.7 V = 3.3 V 




(b) 


7 7 V E 3.3 V 4 

Ic — Ie — 2.75 mA 

R e 1.2 kQ 




(c) 


V c = V cc ~ IcRc = 18 V - (2.75 mA)(2.2 kQ) 

= 11.95 V 




(d) 


V CE = V C -V E = 11.95 V- 3.3 V = 8.65 V 




(e) 


r„ = F[ - f , = i i*v-4v = 
R b R b 330 kQ 




(f) 


I c= 2.75 mA =n416 
I B 24.09 //A 


27. 


(a) 


J _V CC +V EE -V BE _ 6 V + 6 V -0.7 V 

5 i? s +(y? + l)7? £ 330kQ + (121)(1.2kQ) 



= 23.78 juA 

I E = (fi+iy B = (121)(23.78 juA) 

= 2.88 mA 

—Vee + IeRe ~ Ve = 0 

V E = -V E e + IeRe = -6 V + (2.88 mA)(l .2 kQ) 

= -2.54 V 

/X , Vee-Vbe 12 V-0.7 V 

(a) I B = — — — = 

R B +(p + \)R E 9.1 kQ + (120 + 1)15 kQ 

= 6.2 juA 

(b) I c = ph = (120)(6.2 juA) = 0.744 mA 

(c) Vce = Vcc + Vee ~ P(Rc + Re) 

= 16 V + 12 V - (0.744 mA)(27 kQ) 

= 7.91 V 

(d) V c = V cc ~ IcRc = 16 V - (0.744 mA)(12 kQ) = 7.07 V 



41 

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r 8V-0.7V 7.3 V t 

(a) I E = = = 3.32 mA 

2.2 kQ 2.2 kQ 

(b) V c = 10 V-(3.32mA)(1.8kQ) = 10 V- 5.976 

= 4.02 V 

(c) V CE = 10 V + 8 V - (3.32 mA)(2.2 kQ + 1.8 kQ) 

= 18 V- 13.28 V 

= 4.72 V 

(a) j3R E > 10R 2 not satisfied /.Use exact approach: 

Network redrawn to determine the Thevenin equivalent: 



^ o 7 1 + 

'510 kQ ^510 k£2 



Th Thevenin 
18 V 




(b) I c = Ph = (130)(13.95 juA) = 1.81 mA 



510 kQ 



255 kQ 



, 18V+18V A 

/ = = 35.29 //A 

510 kQ + 510 kQ 

E rh = -18 V + (35.29 //A)(510 kQ) 

= 0 V 

= 18 V-0.7 V 

5 255 kQ + (130 + 1)(7.5 kQ) 

= 13.95 juA 



(c) Ve — -18 V + (1.81 mA)(7.5 kQ) 

= -18 V+ 13.58 V 

= -4.42 V 

(d) V CE = 18 V+ 18 V-(1.81 mA)(9.1 kQ + 7.5 kQ) 

= 36 V- 30.05 V = 5.95 V 



r V V C -V RF 8 V-0.7 V t 

(a) I B =-^- = -E = = 13.04 juA 

R b R b 560 kQ 



v cc -v c = 18 V-8 V _ 10V 
R r 3.9 k Q _ 3.9 kQ 



2.56 mA 



/, = 1 56mA 
I B 13.04 //A 



(d) Vce=V c = 8 V 



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h= -*-= 



I„ 2.5 mA 



= 31.25 //A 



p 80 

. V r b _ V CC -V BE _ 12V-0.7V 
I B I B 31.25 /jA 



361.6 kQ 



V cc - V c _V cc -Vce q _ 12 V- 6 V 



2.4 kQ 



2.5 mA 2.5 mA 



R c +R e 



4 -R r +R„ 



10mA 



10 mA: 



20 V 

10 mA => 5 R e = = 2 kQ 

10mA 



400 Q 



R c = 4R E = 1.6 kQ 

y I ( 5 mA . _ _ _ . 

I B = — = =41.67 uA 

P 120 



_ ___ _ 20 V - 0.7 V - 5 mA(0.4 kQ) 19.3-2 V 

Rb ~ V RB'J-B — 

41.67 juA 41.67 juA 

= 415.17 kQ 

Standard values: R E = 390 Q, R c = 1.6 kQ, R B = 430 kQ 



VeJTe 3 V 

I E I c 4 mA 



0.75 kQ 



!' - I ’ V cc -*(Vce „ +V e) 



= 24V-(8 V + 3 V) _ 24V-11 V _ 13 V 
4 mA 4 mA 4 m A 

V B =V E + V be = 3 V + 0.7 V = 3.7 V 

=> 3.7 V = ^- (24 V) 1 2 unknowns! 



3.25 kQ 



7?2 “ 1 “ 



R 2 + 7 ?^ 



use PR E > 107^2 for increased stability 
(1 10)(0.75 kQ) = 10tf 2 
R 2 = 8.25 kQ 
Choose R 2 = 7.5 kQ 



43 

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Substituting in the above equation: 
3?v _ 7.5 kQ(24 V) 

7.5 kQ + R, 



Ri = 41.15 kQ 

Standard values: 



Re = 0.75 kQ, R c = 3.3 kQ, R 2 = 7.5 kQ, R x = 43 kQ 



35 . 



Ve=-Vcc = \( 28V)=5.6V 

R e =^ = = 1.12 kQ (use 1.1 kQ) 

I E 5 mA 

tt' 28 V 

r c = — + V E = +5.6 V= 14 V + 5.6 V= 19.6 V 

2 E 2 

= r cc - Fc = 28 V- 19.6 V= 8.4 V 



Rc = 



_ 



14 V 



I c 5 mA 



= 1.68 kQ (use 1.6 kQ) 



V B = V be + =0.7 V + 5.6 V = 6.3 V 



% 

7?2 R\ 



6.3 V = 



(28 V) 
R 2 + 



(2 unknowns) 



7, = 2mA _, 35J4 
I B 37 //A 
j3R B = 1 07^2 

(135.14)(1.12kQ)= 10(7? 2 ) 

R 2 = 15.14 kQ (use 15 kQ) 

o! • • (15.14 kQ)(28V) 

Substituting: 6.3 V = 

15.14 kQ + ^ 

Solving, Ri = 52.15 kQ (use 51 kQ) 



Standard values: 

Re =1.1 kQ 
R c = 1.6 kQ 
Ri = 51 kQ 
R 2 = 15 kQ 



36. 


r _ 18 V-0.7 V 


= 8.65 mA = 7 


2kQ 2 kQ 


37. 


For current mirror: 






7(3 kQ) = 7(2.4 


kQ) = 7=2 mA 


38. 


I d q — I dss ~ 0 m A 





44 



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39 . 




-9 V 
= / 



40. 



V Z -V B E _ 5.1V-0.7V 
7^ 1.2 kQ 



3.67 mA 




Standard values: 

R b = 43 kQ 

7?c=0.62kQ 



45 

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43. (a) From Fig. 3.23c: 

4 = 2 mA: 4 = 38 ns, 4 = 48 ns, t d = 120 ns, 4=110 ns 
ton = t r + t d = 48 ns + 120 ns = 168 ns 
t Q ff=t s + 4 = 110 ns + 38 ns = 148 ns 

(b) / c = 10 mA: 4 = 12 ns, 4=15 ns, t d = 22 ns, 4 = 120 ns 
ton = 4 + t d = 1 5 ns + 22 ns = 37 ns 
4 f f = 4 + 4 = 120 ns + 12 ns = 132 ns 
The turn-on time has dropped dramatically 
168 ns:37 ns = 4.54:1 

while the turn-off time is only slightly smaller 
148 ns: 132 ns = 1.12:1 




44. (a) Open-circuit in the base circuit 

Bad connection of emitter terminal 
Damaged transistor 

(b) Shorted base-emitter junction 
Open at collector terminal 

(c) Open-circuit in base circuit 
Open transistor 

45. (a) The base voltage of 9.4 V reveals that the 1 8 kQ resistor is not making contact with the 

base terminal of the transistor. 

If operating properly: 



18kQ(16 V) 
18 kQ + 91 kQ 



2.64 V vs. 9.4 V 



As an emitter feedback bias circuit: 



J _ Vgc-V,E 16 V-0.7 V 

B R,+ (p + \)R E 91 kQ + (100 + l)1.2 kQ 
= 72.1 //A 

V B = V cc ~ MR i) = 16 V - (72.1 //A)(91 k^) 

= 9.4 V 



46 



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(b) 



Since V E > V B the transistor should be “off’ 



With I B = 0 juA, V B = 



18kQ(16 V) 
18kQ + 91kQ 



= 2.64 V 



.'. Assume base circuit “open” 

The 4 V at the emitter is the voltage that would exist if the transistor were shorted 
collector to emitter. 



1.2 kQ(16 V) =4y 
1.2 kD + 3.6 kQ 



46. (a) R b t, I B i, I c i, Vet 

(b) pi, id 

(c) Unchanged, I c not a function of p 

(d) kcok Ici' 

(e) pi,ld,V Rc i,V RE i,V C E t 



47. 



48. 



(a) 



(b) 

(c) 

(d) 

(e) 

(a) 

(b) 

(c) 

(d) 

(e) 



h = 



E -V 

^ Th v BE 



E -V 

^ Th v BE 



R Th + (P + R Th + P R E 
Ic = Ph= P 



E -V 

_ Th y BE 

R Th + P R E 



E -V 

^ Th Y BE 



pT[ 

P 



+ Ere 



As pt, ^-i,I c t, V Rc t 

Vc=V cc - V Rc 

and Vc'l 



R 2 = open, Ib \ , / c t 

Vce = Vcc~ Ic(Rc + Re) 
and V C e ^ 

Vcc^i Vb'I’, Ve'I', Ic\ 

I B = o juA , I c = Iceo and / c (i?c + A £ ) negligible 
with Kce = fee = 20 V 

Base-emitter junction = short 7 5 T but transistor action lost and I c = 0 mA with 

Vce=V cc = 20 V 



7?^ open, I B — 0 //A, Ic — Iceo = 0 mA 
and Uc= U cc =18 V 

A/ct, Fj, F,t, F C£ I 

Drop to a relatively low voltage = 0.06 V 
Open in the base circuit 



47 

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49. 



22.16 juA 



j _ Vcc-Vbe _ 12 V-0.7V _ 11.3 V 
B R b 510 kQ 510 kQ 

I c = Ph = (100)(22.16 juA) = 2.216 mA 
V c = ~Vcc + IcRc = -12 V + (2.216 mA)(3.3 kQ) 
= -4.69 V 
V CE = Vc = - 4.69 V 



50. ^>107? 2 

(220)(0.75 kQ)> 10(16 kQ) 
1 65 kQ >160 kQ (checks) 
Use approximate approach: 



16 kQ(-22 V) 
16 kQ + 82 kQ 



= -3.59 V 



V e =V b + 0.7 V = -3.59 V + 0.7 V = -2.89 V 
I c = h= Ve/Re = 2.89/0.75 kQ = 3.85 mA 



/ c _ 3.85 mA 

/ B — — 

p 220 



17.5 juA 



y c V cc “I" IcRc 

= -22 V + (3.85 mA)(2.2 kQ) 

= -13.53 V 



51. 



T - V-Vbe 

1e 



8 V - 0.7 V 7.3 V 



R* 



3.3 kQ 



3.3 kQ 



= 2.212 mA 



V c = - V cc + IcRc = -12 V + (2.212 mA)(3.9 kQ) 

= -3.37 V 

52. (a) S{I co ) = P+ 1=91 



(b) S(V HE )= V = - 90 



R r 470 kQ 



= -1.92 xlO S 



(c) S(p)=A= — = 32.56 X 10" 6 A 

P x 90 

(d) M c = SdcoWco + S( V be )A V be + S(J3) Ap 

= (91)(10 //A - 0.2 //A) + (-1.92 x 10“ 4 S)(0.5 V - 0.7 V) + (32.56 x 10“ 6 A)(1 12.5 - 90) 
= (91)(9.8 juA) + (1.92 x lO^SXOJ V) + (32.56 x 10“ 6 A)(22.5) 

= 8.92 x 10“ 4 A + 0.384 x lO^A + 7.326 x lO^A 
= 16.63 x 10“ 4 A 
= 1.66 mA 



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53. 



For the emitter-bias: 



(a) S(Ico) = (fi+l) 



(1 + R b / r e) = (! oo + 1) 
(P + \) + R B /R E 



(1 + 510 kQ/1.5 kQ) 
(100 + 1) + 5 10 kQ/1.5 kQ 



= 78.1 



(b) S(V BE ) = 



~P 



-100 



R b + (/? + l)R E 5 10 kQ + (100 + 1)1 .5 kQ 



= -1.512 x 1(TS 



/ Q ( 1 + V^) = 2.92 mA(l+ 340) 

Cj {P p x {\ + P 2 +R b IR e ) 100(1 + 125 + 340) 

= 21.37 x 10“ 6 A 



(d) Ale = S(I C oWco + S(V be )AV be + S(J3)AJ3 

= (78.1)(9.8 juA) + (-1.512 x 10“ 14 S)(-0.2 V) + (21.37 x 10“ 6 A)(25) 
= 0.7654 mA + 0.0302 mA + 0.5343 mA 

= 1.33 mA 



54. (a) R T h = 62 kQ || 9.1 kQ = 7.94 kQ 

Wco) = m 1) 1 + = (80 + D- (1 + 734 kQ/0 - 68 kQ) 



(J3 + 1) + R T JR E 
(81X1 + U-68) 



(80 + 1) + 7.94 kQ / 0.68 kQ 



11.08 



(b) S(V BE ) = 



81 + 11.68 

~P 

R n +(J3 + 1 )R e 7.94 kQ + (81)(0.68 kQ) 



-80 



-80 



7.94 kQ + 55.08 kQ 



= -1.27 x 10" 3 S 



+(! + RtJRe) 1.71 mA(l +7.94 kQ/ 0.68 kQ) 

C) ( 3 (\ + p 2 + R n /R e )~ 80(1 + 1 00 + 7.94 kQ / 0.68 kQ) 

= 171mA(12.68)^ 41xl() - 6A 
80(112.68) 



(d) M c = S(I co )Mco + S( V BE ) A V BE + S(J3)AJ3 

= (1 1.08)(10 juA - 0.2 juA) + (-1.27 x 10“ 3 S)(0.5 V - 0.7 V) + (2.41 x 10“ 6 A)(100 - 80) 
= (1 1.08)(9.8 /jA) + (-1.27 x 10“ 3 S)(-0.2 V) + (2.41 x 10+0(20) 

= 1.09 x 10+ + 2.54 x 10+ + 0.482 x 10+ 

= 4.11 x 10+. = 0.411 mA 



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55. For collector-feedback bias : 

(a) S(Ico) = (/?+!) n + /?fl//?c) 



= (197.32) 



(J3 + 1) + R B /R C 
1 + 143.59 



= (196.32 + 1) 



(1 + 560 kQ/3.9 kQ) 
(196.32 + 1) + 560 kQ/3.9 kQ 



(197.32 + 143.59) 



= 83.69 



(b) S(V BE ) = 



~P 



-196.32 



R b + (/? + 1 )R C 560 kQ + (196.32 + 1)3.9 kQ 



= -1.477 x 10“ 4 S 



(c) S(j3) = 



I a (Rb +R C ) 



2.56 mA(560 kQ + 3.9 kQ) 



p x (R B +R c (p 2 + 1)) 196.32(560 kQ + 3.9 kQ(245.4 + l)) 



= 4.83 x 10 A 



(d) AIc = S(IcoWco + S(V BE )AV BE + S({J)Aj3 

= (83.69)(9.8 juA) + (-1.477 x 10“ 4 S)(-0.2 V) + (4.83 x 10“ 6 A)(49.1) 
= 8.20 x 10“ 4 A + 0.295 x 10“ 4 A + 2.372 x 10“ 4 A 
= 10.867 x 10“ 4 A= 1.087 mA 



Type 


S(Ico) 


S(V B e) 


S(P) 


Collector feedback 


83.69 


-1.477 x 10“ 4 S 


4.83 x 10“ 6 A 


Emitter-bias 


78.1 


-1.512 x 10“ 4 S 


21.37 x 10“ 6 A 


Voltage-divider 


11.08 


-12.7 x lO^S 


2.41 x 10“ 6 A 


Fixed-bias 


91 


-1.92 x lO^S 


32.56 x 10“ 6 A 



S(I C oY Considerably less for the voltage-divider configuration compared to the other three. 
S(V BE )\ The voltage-divider configuration is more sensitive than the other three (which have 
similar levels of sensitivity). 

S(f3): The voltage-divider configuration is the least sensitive with the fixed-bias 
configuration very sensitive. 

In general, the voltage-divider configuration is the least sensitive with the fixed-bias the most 
sensitive. 



57. (a) Fixed-bias: 

S(Ico) = 91, AI C = 0.892 mA 
S(V BE ) = -1.92 X 10“ 4 S, Ale = 0.0384 mA 
S(J3 ) = 32.56 x K+A, A/ c = 0.7326 mA 
(b) Voltage-divider bias: 

S(I C0 ) = 1 1 .08, Ale = 0. 1090 mA 
S(V BE ) = -1.27 x 10“ 3 S, Ale = 0.2540 mA 
S(J3) = 2.41 x 10“ 6 A, Ale = 0.0482 mA 



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(c) For the fixed-bias configuration there is a strong sensitivity to changes in I C o and / 3 and 
less to changes in Vbe- 

For the voltage-divider configuration the opposite occurs with a high sensitivity to 
changes in V BE and less to changes in I C o and /?. 

In total the voltage-divider configuration is considerably more stable than the fixed-bias 
configuration. 



51 

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Chapter 5 



(a) If the dc power supply is set to zero volts, the amplification will be zero. 

(b) Too low a dc level will result in a clipped output waveform. 

(c) P 0 = I 2 R = (5 mA) 2 2.2 kQ = 55 mW 

Pi = V CC I= (18 V)(3.8 mA) = 68.4 mW 



7 / : 



P Q { ac) 55 mW 



/>(dc) 68.4 mW 



= 0.804 => 80.4% 



3. x c =— — = = 15.92 Q 

2nfC 2n(\ kHz)(l 0 //F) 

/= 100 kHz: x c = 0.159 Q 
Yes, better at 100 kHz 



= V i _ 10 mV 
* 7 ~ 0.5 mA 

= 20 Q (=r e ) 

V 0 = I c R l 
= aI c Ri 

= (0.98)(0.5 mA)(1.2kQ) 

= 0.588 V 

K = 0588V 
V. 10 mV 

= 58.8 

(d) Z 0 = oo Q 

/ al 

(e) 4 = — = — = «=0.98 

i, K 

(f) h = i e -i c 

= 0.5 mA - 0.49 mA 

= 10 juA 



5. (a) 

(b) 



52 

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, , v t 48 mV 1__ 

(a) r e = — = = 15 Q 

I 3.2 mA 



(b) Z t = r e = 15 Q 

(c) I c =ctf e = (0.99)(3.2 mA) = 3.168 mA 

(d) V 0 = I c R l = (3.168 mA)(2.2 kQ) = 6.97 V 



V 6.97 V 

(e) A v = —= = 145.21 

V. 48 mV 



(f) I b = (1 - d)I e = (1 - 0.99)4 = (0.01)(3.2 mA) 

= 32 juA 

n , , 26 mV 26 mV 

7. (a) r e = = = 13 Q 

I E (dc) 2 mA 

Zi = pr e = (80)(13 Q) 

= 1.04 kQ 






/ 



p p p + i 

2 mA 



/?+! 



81 



= 24.69 ,uA 



(c) 4=^=4 



/,.= 



fpiPh) 
r + R, 



J r + R, 

At= — 



•M 



40 kQ 



40 kQ + 1.2 kQ 

= 77.67 



r +7?, 



(80) 



•/? 



(d) A v =- 



R, r 

L || O 



1.2 kQ 40 kQ 



13 Q 



1.165 kQ 
13 Q 

= -89.6 



53 

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(a) Z, = /iv = (140 )r,= 1200 

= 1^2® = 8.571 Q 
140 

„ . , V t 30 mV . 

(b) 4= z=II^ =25M 



(c) I c = Ph = (140>(25 //A) = 3.5 mA 

_ _rj_c — _ (50 kQ)(3.5 mA) _ 3 321 mA 
r o +R L 50kQ + 2.7kQ 

A,- i. = 3321mA „ 

/. 25 //A 



(e) ^ = ^ = X132.84)(^ 
K Z, 1.2 kQ 



K - F 

(a) r e : I B = cc BE 

Rr 



= -298.89 

12 V-0.7 V 



220 kQ 



= 51.36 juA 



I E = (/?+ 1)A = (60 + 1)(51.36 //A) 
= 3.13 mA 
26 mV 26 mV 



= 8.31 Q 



I E 3.13 mA 
Z* = R b || pr e = 220 kQ || (60X8.31 Q) = 220 kQ || 498.6 Q 

= 497.47 Q 

r Q > 107?c Z 0 = Rc = 2.2 kQ 



R —2.2 kQ 

(b) A v = — £- = — = -264.74 

r 8.31 Q 



(c) Zi = 497.47 Q (the same) 

Z 0 = r 0 \\R c = 20 kQ || 2.2 kQ 

= 1.98 kQ 



(d) 



A v = 



~Rr 



At = -AyZJRc 



-1.98 kQ 
8.31 Q 



= -238.27 



= -(-238.27)(497.47 Q)/2.2 kQ 

= 53.88 



54 

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26 mV 



R c _ 4.7 kQ = 

T” (-200) " 

26 mV 26 mV 



23.5 Q 



23.5 Q 



1.106 mA 



L. 106 mA 



12.15 juA 



V -V 

v CC v BE 



V CC — IbRb + V BE 



(12.15 juA){\ MQ) + 0.7 V 
= 12.15 V + 0.7V 

12.85 V 



(a) /^^k = 10V -°- 7V ^3.85 M 
R b 390 kQ 

I E = (/? + 1)A = (101)(23.85 //A) = 2.41 mA 
26 mV 26 mV 



2.41 mA 



10.79 Q 



Ic = Ph = (100)(23.85 juA) = 2.38 mA 



(b) Z, = R B \\pr e = 390 kQ || (100)( 10.79 Q) = 390 kQ || 1.08 kQ 

= 1.08 kQ 

r Q > 107?c • * Z 0 = R c = 4.3 kQ 



^ = -43kD 
r 10.79 Q 



= -398.52 



R r \\r 

(d) A v = — ^ 
r 



(4.3 kQ) 11(30 kQ) 3.76 kQ 



10.79 Q 



1 0.79 Q 



-348.47 



(a) Tzst PR e > 107^2 
? 

(100)(1.2 kQ) > 10(4.7 kQ) 

120 kQ > 47 kQ (satisfied) 



Use approximate approach: 

Rf cc _ 4.7kiX16V) _, 7 , lv 

R } +R 2 39 kQ + 4.7 kQ 

V E = V b -V be = 1.721 V- 0.7 V= 1.021 V 

r V F 1.021V . 

I E = — = = 0.8507 mA 

R„ 1.2 kQ 



26 mV 



26 mV 
0.8507 mA 



30.56 Q 



www.elsolucionario.net 




(b) Zi = R\ || R 2 || pr e 

= 4.7 kQ || 39 kQ || (100)(30.56 Q) 

= 1.768 kQ 

r 0 > 10i?c Z 0 = Rc= 3.9 kQ 



(c) A v = 

r e 



3.9 kQ 
30.56 Q 



= -127.6 



(d) r Q = 25 kQ 

(b) Zi (unchanged) = 1.768 kQ 

4 = R c || r„ = 3.9 kQ || 25 kQ = 3.37 kQ 

A _ (7? c ||r 0 ) (3.9 kQ)|| (25 kQ) 3.37 kQ 

Cj v r e 30.56 Q 30.56 Q 

= -110.28 (vs. -127.6) 



13. J3R e > 10 R 2 

(100)(1 kQ)> 10(5.6 kQ) 

100 kQ > 56 kQ (checks!) &r Q > 10 R c 
Use approximate approach: 



A v = 

r e 

26 mV 



A 

A 



3.3 kQ 
" -160 



= 20.625 Q 



>*E = 



26 mV 26 mV 



20.625 Q 



= 1.261 mA 






I E = V E = I e R e = (1.261 mA)(l kQ) = 1.261 V 

R e 

V B = V BE + Fe = 0.7 V+ 1.261 V= 1.961 V 
5.6k DK CC , 196[V 

5.6 kQ + 82 kQ 

5.6 kQ F cc = (1.961 V)(87.6 kQ) 

Fee = 30.68 V 



14. Test f3R E > 107^2 
? 

(180)(2.2kQ) > 10(56 kQ) 

396 kQ <560 kQ (not satisfied) 

Use exact analysis: 

= 56 kQ || 220 kQ = 44.64 kQ 

= 56knt20V) - 4.058 V 

220 kQ + 56 kQ 

E Th~ V BE _ 4.058 V -0.7 V 

R Th + (fi + 1 )R e ~ 44.64 kQ + (1 8 1)(2.2 kQ) 



(a) R n = 

Eth = 

h = 



56 

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= 7.58 juA 

Ie = (P+1)Ib = (181)(7.58 juA) 
= 1.372 mA 



26 mV 26 mV 



1.372 mA 



= 18.95 Q 



V E = IeRe = (1.372 mA)(2.2 kQ) = 3.02 V 
Vb=V e + V be = 3.02 V + 0.7 V 
= 3.72 V 

Vc = Vcc ~ PRc 

= 20 V - pI B R c = 20 V - (180)(7.58 //A)(6.8 kQ) 

= 10.72 V 

Z i = R X \\R 2 \\Pr e 

= 56 kQ || 220 kQ || (180)(18.95 kQ) 

= 44.64 kQ || 3.41 kQ 

= 3.17 kQ 



R r \\r 

r 0 <\ 0R C .-. A v = 

(6.8 kQ)|| (50 kQ) 

18.95 Q 

= -315.88 



Ib = 



V -V 

v CC v BE 



20 V-0.7 V 



R B +{fi + \)R E 
19.3 V 



390 kQ + (141)(1.2 kQ) 
= 34.51 /uA 



559.2 kQ 

I E = (P+ 1)/ 5 = (140+ 1)(34.51 juA) = 4.866 mA 
26 mV 26 mV 



r e = 



4.866 mA 



= 5.34 Q 



Zb~ pr e + (P+ 1 )Re 

= (140)(5.34 kQ) + (140 + 1)(1.2 kQ) = 747.6 Q + 169.9 kQ 

= 169.95 kQ 

Zi = R b || = 390 kQ || 169.95 kQ = 118.37 kQ 

Z 0 = R C = 2.2 kQ 

A _ PR C _ (140X2.2 kQ) _ 1S1 

v Z b 169.95 kQ 



Zb ~ pr e + 



(P + 1) + Rg / r o 
l + (R c +R E )lr o 



= 747.6 Q 



(141) + 2.2 kQ/20 kQ 
1 + (3.4 kQ) / 20 kQ 



57 

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= 747.6 Q + 144.72 kQ 
= 145.47 kQ 



Z* = R b || Z 6 = 390 kQ || 145.47 kQ = 105.95 kQ 
Z Q = R c = 2.2 kQ (any level of r Q ) 




pR c 



1 + 



Rr 



1 + ^ 



-(140)(2.2 kQ) 


r + 5.34^1 


2.2 kQ 


145.47 kQ 


/2tfkQ 


i 

20 kQ 



2.2 kQ 
20 kQ 



-2.117 + 0.11 

ul 



= -1.81 



16. Even though the condition r Q > 107? c is not met it is sufficiently close to permit the use of the 
approximate approach. 






*c_ 

10 



PR C _ R c = _ j q 



m E 

$.2 kQ 
10 



R* 



= 0.82 kQ 



, 26 mV 26 mV _ A 

I E = = = 6.842 mA 

r e 3.8 Q 

V E = IeRe = (6.842 mA)(0.82 kQ) = 5.61 V 
V B = V E + V BE = 5.61 V + 0.7 V = 6.31 V 



Ib = 



V D 



and R b = 



6.842 mA 
121 



Vcc-Vb 



= 56.55 /jA 



20 V-6.31 V 
56.55 juA 



= 242.09 kQ 



17. (a) dc analysis the same 

.'. r e = 5.34 Q (as in #15) 

(b) Z, = R b || = R b || pr e = 390 kQ || (140)(5.34 Q) = 746.17 Q vs. 1 18.37 kQ in #15 

Z 0 = R C = 2.2 kQ (as in #15) 

(c) A v = = — kQ =-411.99 vs -1.81 in #15 

r e 5.34 Q 

(d) Z, = 746.17 Q vs. 105.95 kQ for #15 

Z 0 = R c || r„ = 2.2 kQ || 20 kQ = 1.98 kQ vs. 2.2 kQ in #15 



58 



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A v = — 



Rr 



1.98 kQ 



= -370.79 vs. -1.81 in #15 



r e 5.34 Q 
Significant difference in the results forZ v . 



18 . 



v - V 

(a) I B = ^ ^ 



Rb + iP + 1)Re 

22 V-0.7 V 



21.3 V 



330 kQ + (8 1)(1 .2 kQ + 0.47 kQ) 465.27 kQ 
= 45.78 juA 

I E = (fi+ 1)/^ = (81)(45.78 juA) = 3.71 mA 



r e = 



26 mV 26 mV 



I E 3.71 mA 



= 7Q 



(b) r Q < 1 0(Rc + R e ) 

{ p + X) + Rc /r o 



:.Z b - (3r e + 



l + (7? c +R e )I r o 



R* 



= (80)(7 Q) + 



(81) + 5.6 kQ/40 kQ 
1 + 6.8 kQ/40kQ 



= 560 Q + 



81 + 0.14 
1 + 0.17 



1.2 kQ 



(note that (/?+ 1) = 81 » Rc^o = 0.14) 

= 560 Q + [81.14 /1.17J1.2 kQ = 560 Q + 83.22 kQ 

= 83.78 kQ 



Zi = R b || Z* = 330 kQ || 83.78 kQ = 66.82 kQ 



-PRc 



i + - 



A v = 



Rr 



1 + ^ 



~(80)(5.6 kQ) ( 7 Vy \ | 5.6 kQ 

== 83.78 kQ [ + 4j/knj40kQ 

1 + 5.6 kQ/40 kQ 
_ -(5.35) + 0.14 
1 + 0.14 
= -4.57 



19 . 



V cc~ V be _ 16 V-0.7 V _ 15.3 V 

R b + (fi + \)R E ” 270 kQ + (1 1 1)(2.7 kQ) “ 569.7 kQ 



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= 26.86 juA 

Ie = (J3+ 1)4 = (110 + 1)(26.86//A) 
= 2.98 mA 



26 mV 26 mV 



= 8.72 Q 



I E 2.98 mA 
pr e = (1 10)(8.72 Q) = 959.2 Q 



(b) Z b = pr e + (J3+ 1)Re 

= 959.2 Q + (lll)(2.7 kQ) 

= 300.66 kQ 

Z z = R b || Z 6 = 270 kQ || 300.66 kQ 
= 142.25 kQ 

Z G = R e || r e = 2.7 kQ || 8.72 Q = 8.69 Q 



(c) 



A v = 



Rr 



R E+ r e 



2.7 kQ 

2.7 kQ + 8.69 Q 



0.997 



20 . 



(a) 



Vce-Vee _ 8 V- 0.7 V 

R b +(P + 1 )R e 390 kQ + (121)5.6 kQ 



h ~ 
r e = 



1 )I B = (121)(6.84 /uA) = 0.828 mA 
26mV = 31.4 Q 



I E 0.828 mA 



r a < 10 R e : 



Zb ~ fir e + 



Qg + i )Re 
1 + R E /r 0 



= (120X31.40)+ (l 21 )(5.6kn) 

1 + 5.6 kQ/40kQ 
= 3.77 kQ + 594.39 kQ 



= 598.16 kQ 

Zi = R b || Z 6 = 390 kQ || 598.16 kQ 

= 236.1 kQ 



Z 0 = R e || r e 

= 5.6 kQ || 31.4 Q 

= 31.2 Q 



6.84 juA 



(b) 



_ Qg + 1)^/Z, 

1 + R E lr 0 

= (121)(5.6kQ)/598.16kQ 
1 + 5.6 kQ/40 kQ 

= 0.994 



(c) A v =^= 0.994 
Vt 

V 0 =A v Vt = (0.994)(1 mV) = 0.994 mV 



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21. (a) 



? 

Test J3R-E — 107^2 
(200)(2 kQ) > 10(8.2 kQ) 

400 kQ > 82 kQ (checks)! 



Use approximate approach: 
_ 8.2 kQ(20 V) 

v B - 



= 2.5545 V 



8.2 kQ + 56 kQ 
V E = V b - V BE = 2.5545 V - 0.7 V = 1.855 V 



. _ V E 1.855 V 
b — - 



R„ 



b = 



2 kQ 
0.927 mA 



0.927 mA 



= 4.61 juA 



(/? + 1) (200+ 1) 
b = Pb = (200)(4.61 /uA) = 0.922 mA 



(b) r e 



26 mV 

h 



26 mV 
0.927 mA 



= 28.05 Q 



(c) Z b = Pr e + (P+\)R E 

= (200)(28.05 Q) + (200 + 1)2 kQ 
= 5.61 kQ + 402 kQ = 407.61 kQ 
Zi = 56 kQ || 8.2 kQ || 407.61 kQ 
= 7.15 kQ || 407.61 kQ 
= 7.03 kQ 

4 = R E \\r e = 2 kQ || 28.05 Q = 27.66 Q 

(d) A v = Re = = 0.986 

R E +r e 2kD + 28.05 Q 



22 . 



.. r V EE -V BE 6V- 0.7V A 

(a) I E = — — = — — = 0.779 mA 



R „ 



6.8 kQ 



26 mV 26 mV „ ^ 

r e = = = 33.38 Q 

I E 0.779 mA 



(b) Zi = R e || r e = 6.8 kQ || 33.38 Q 

= 33.22 Q 

Z n = R c = 4.7 kQ 



(c) ly _ aRc_ (0.998X4.7 kQ) 



33.38 Q 



= 140.52 



23. 



6 75 

cr= -J— = — =0.9868 
P + 1 76 



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= 1.1 mA 



V PE _-V Ri 

R* 



5V-0.7V 4.3 V 



3.9 kQ 3.9 kQ 



26 mV 26 mV ^ co 0 

r e = = = 23.58 12 

I E 1.1mA 

R c _ (0.9868)(3.9 kQ) 

r ~ 23.58 Q 



24 . 



V -V 

(a) I B =^ ^ 



12 V- 0.7 V 



R f + PR C 220 kQ + 120(3.9 kQ) 
= 16.42 /uA 

I E = (P+ 1 )Ib = (120 + 1)(16.42 /uA) 

= 1.987 mA 



26 mV 26 mV 



I E 1.987 mA 



= 13.08 Q 



R „ 



(b) Zi = Pr e \\^f- 

Kl 



Need 4,! 



-3.9 kQ 
13.08 Q 



Zi = (120)(13.08 Q) | 



= -298 



220 kQ 
298 



= 1.5696 kQ || 738 Q 

= 501.98 Q 

Z Q = R c || Rf = 3.9 kQ || 220 kQ 

= 3.83 kQ 

(c) From above, A v = -298 



25. 



A v = -A- =-160 



R c = 1 60(r,) = 160(10 Q) = 1.6 kQ 



A - 

R e + PR C 



19 => 19 = 



200 R f 

R f + 200(1.6 kQ) 



19R f +3800R c = 200R f 

_ _ 38007? c _ 3800(1.6 kQ) 

R i 

181 181 



= 33.59 kQ 



J Vcc-Vbe 

R f + PR C 

Ib(Rf + PRc) = Vcc ~ V be 



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and Vcc ~ V be + Ib(Rf + pRc) 



with //. = 



26 mV 26 mV 



Ib = 



ion 

2.6 mA 



= 2.6 mA 



P + 1 200 + 1 



= 12.94 //A 



Fee — ksE + Ib(Pf + )® c ) 

= 0.7 V + (12.94 //A)(33.59 kQ + (200)(1.6 kQ)) 

= 5.28 V 




(a) 4,: V^hPre + iP+iytRE 

i 0 + r = i c = pi b 
but ii = r + a 

and T = 7 - A 

Substituting, A + (/* - A) = pi h 
and 4 = 00+ 1)4 - Ii 

Assuming (/?+ 1)4 » h 
I 0 = (J3+ 1)4 

andK 0 = -/^ c = -(^+l)/^c 



Therefore, — = 



~(yg + l)/^ c 
I b pr e +{(l + \)I b R E 



and = 






/+*/<• + ftfb R E 

R c ^ 

r e + 



(b) V, = pl h (r e + R E ) 
For r e «; 
Vt^PhRE 



Now /,=/' + I b 
V,—V„ 



Rr 



- + L 



63 

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Since V„ » V, 




or I i, = /, + 



K 

r f 



and 



but 

and 



V t = pi h R E 

V i = pR E I i + P^-R e 



V 0 =A v V, 



Vi = pRsIi + 



Rf 



or 



Vt - 



Vi 



MMl 

Rf 

AM, 

R „ 



pRfl, 

[PR E Vi 



so z - v > - P r b PMe 

‘ h x _AMe_ R F +p(-A r )R E 
Rf 



Zt = 



Vi 

I, 



=*\\y 



where x = J3R E and y = 74/p4 v | 



with Zi = 






X7 og^x^/KD 

x + y PR E + R F l\A v \ 

P R l R f 

PR, |.l, + R, 



Z a : SetJ^ = 0 




Vi = I b pr e + (P+ 1 )I„R e 
Vi = pI b {r e + R E ) = 0 

since J3,r e + R E ^0 4 = 0 and fil b = 0 



64 

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V V 

.*. I Q = -^ + ^ = V 
R c R f 

and Z 0 = Y = -^~ 



R c R F 



R c R F 



Rc^F _ n || n 



(c) 1.83 



R* 



1.2 kQ 



Z = 



PReRp 



(90)(1.2 kQ)(120 kQ) 



| A v | + (90)(1 .2 kQ)(l .83) + 120 kQ 

= 40.8 kQ 
Z 0 = R c \\Rf 
= 2.2 kQ || 120 kQ 

= 2.16 kQ 



27. 



v -V 

(a) I B =-^~ ^ 



9 V-0.7V 



R f + pR c (39 kQ + 22 kQ) + (80)(1 .8 kQ) 

8.3 V 8.3 V .... A 

= = = 40.49 //A 

61 kQ + 144 kQ 205 kQ 

I E = (J3 + 1)1 b = (80 + 1)(40.49 //A) = 3.28 mA 
26 mV 26 mV 



I E 3.28 mA 



= 7.93 Q 



Zi=R Pl II Pr e 

= 39 kQ || (80)(7.93 Q) = 39 kQ || 634.4 Q = 0.62 kQ 
Z c = R c || Rf = 1.8 kQ || 22 kQ = 1.66 kQ 



, Ul „ _ -R’ _ -^c||^ 2 _ 1.8 kQ 1 22 kQ 
(b) A v — — — 

r r 7.93 Q 



-1.664 kQ 
7.93 Q 



= -209.82 



28. 

29. 



At = 0=60 
At* 0=100 



30. At = -AyZJRc = -(-127.6X1.768 kQ)/3.9 kQ = 57.85 

31. (c) 4=-***-= (140X390 kQ) 

R B +Z b 390 kQ + 0.746 kQ 

(d) A, = -A v A = -(-370.79)(746.17 Q)/2.2 kQ 

Rc 

= 125.76 



65 

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32. 



A t = -A v ZJR e = -(0.986)(7.03 kQ)/2 kQ = -3.47 



/ al 

33. Ai= = ^ = a= 0.9868= 1 
L I 



34. 

35. 



4- = -A v ZJR c = -(-298X501.98 Q)/3.9 kQ = 38.37 
A t = -A^^ - ( - 2Q9 - 82Xa62kQ) = 72.27 






1.8 kQ 



36. 



T V CC -V BF 18 V -0.7 V . 

(a) 7^= cc ^ ** = — — =25.44 //A 






680 kQ 



I E = (J3+ l)I B = (100 + 1X25.44 juA) 
= 2.57 mA 
26 mV 



= 10.1 16 Q 



2.57 mA 

A v = = — 3 3 kQ = -326.22 

VAa r e 10.116 Q 

Zt = R b || pr e = 680 kQ || (100X10.1 16 Q) 
= 680 kQ || 1,011.6 Q 

= 1.01 kQ 

Z 0 — R c = 3.3 kQ 



(b) - 

( c ) < = 



R, 



Rl+R, 



~^V nl 



4.7 kQ 



4.7 kQ + 3.3 kQ 



(-326.22) 



= -191.65 



(d) A Il = | l = -(-191.65)- (101 kQ) 



4.7 kQ 



41.18 



(e) A - -M (R < II R L ) -939 kQ) 

u Vt v t /,, ( fir , ) /tlf0(10.116Q) 



= -191.98 

Z, = R b || pr e = 1.01 kQ 

Il = =41.257* 

R c +R l 



h = 



RrI ‘ = 0.99857, 



R B +Pr e 



A = -^ = -L = - 



h h h h 



(41.25)(0.9985) 



= 41.19 

Z 0 = R C = 3.3 kQ 



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37. 



(a) 



A v =-326.22 

V NL 

A A 

A v = — A v 

L nl 

R l = 4.7 kQ: A Vl = 

R l = 2.2 kQ: ^ = 

R l = 0.5 kQ: ^ = 
As Rl'I', A Vl >1 



4.7 kQ 

4.7 kQ + 3.3 kQ 
2.2 kQ 

2.2 kQ + 3.3 kQ 
0.5 kQ 

0.5 kQ + 2.3 kQ 



(-326.22) 

(-326.22) 

(-326.22) 



(b) No change for Z h Z a , and A v ! 



38. 



(a) 



v - V 

t v CC v BE 





R-r 



12V-0.7 V 
1 MQ 



= 11.3 juA 



Ie = (J3+ 1)/ 5 = (181)(11.3 juA) = 2.045 mA 
26 mV 26 mV 



r e = 



A =-^ = 



2.045 mA 
3 kQ 
12.71 Q 



= 12.71 Q 



= -236 



Zi = R B \\pr e =\ MQ || (180)(12.71 Q) = 1 MQ 
= 2.283 kQ 

Z 0 = R( = 3 kQ 



(b) - 

(c) No-load: A v = A = -236 

v 7 '’nl 



(d) 



Z, 2.283 kQ(-236) 

_ Z t + R s Av '" ~ 2.283 kQ + 0.6 kQ 

= -186.9 



(e) V 0 = -l„R c = -phRc 
Vi = I b pr e 

V, Phr e r e 12.71 Q 

V V v. 

A - — — — • — 

v * K v t v s 

v= OMQ 2.288 kQ(F,) 

‘ (1 MQ |yffr e ) + 2.288 kQ + 0.6 kQ 

A v = (-236)(0.792) 

= -186.9 (same results) 



-191.65 

-130.49 

-42.92 



2.288 kQ 



= 0.792 V, 



67 

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(f) No change! 



< g , 



Z. + R c 
Rst A ^ 



2.283 kQ + lkQ 



39 . 



(h) No change! 

/x _ V cc - V BE 24V-0.7V ^ . 

(a) I B = — — = =41.61 uA 



500 kQ 
<41.( 

= 7.715 Q 



A=(/?+ l)/g = (80 + 1)(41 .61 //A) = 3.37 mA 
26 mV 26 mV 



I E 3.37 mA 

. /?, 4.3 kQ 

4 = — A = = -557.36 

VjVL r e 7.715 Q 

Z z = || pr e = 560 kQ || (80)(7.715 Q) 

= 560 kQ || 617.2 Q 

= 616.52 Q 

Z 0 = R c = 4.3 kQ 



(b) - 



V R 

(C) A v =^ = 

1 V t r l +r q 

= -214.98 

A -K-KA 

A ‘~v~v. V 

s i s 

zv 

Vi= — 



A = 



2.7 kQ(-557.36) 
2.7 kQ + 4.3 kQ 



616.52 Q V 



Z ? + R s 616.52 Q + lkQ 
A Vs = (-214.98X0.381) 

= -81.91 



= 0.381 V s 



(d) A= -A 



f R s +Z^ 

V R L J 



= -(-81.91) 

49.04 



lkQ + 616.52 Q 
2.7 kQ 



V R 

(e) A v =^ = 

Vi r l + R < 



_ 5.6 kQ(-557.36) 
VjVi 5.6 kQ + 4.3 kQ 



— the same = 0.381 

A =-^A = (-315.27X0.381) = -120.12 

v s y V 



As^T, ^ T 



68 



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(f) A Vl the same = -214.98 

v, z , 616.52 Q = 0 552 

V s Z i + R s 616.52 Q + 0.5 kQ 

A Vs =y-y = (-214.98X0.552) = -118.67 

As R s i, 4, t 

(g) No change! 

(a) Exact analysis: 

= lO^V) .3.048 V 
R t +R 2 cc 68 kQ + 16 kQ 

Rn = R\ II Ri = 68 kQ || 16 kQ = 12.95 kQ 



F -V 

^ Th v BE 



3.048 V- 0.7 V 



R Th + (j3 + l)R E 12.95 kQ + (101)(0.75 kQ) 
= 26.47 juA 

/* = (/?+ 1)/ 5 = (101)(26.47 //A) 

= 2.673 mA 
26 mV 26 mV 



, -Rc 

A, = — - 



2.673 mA 
2,2 kQ 
9.726 Q 



9.726 Q 



-226.2 



Zf = 68 kQ || 16 kQ || J3r e 
= 12.95 kQ || (100)(9.726 Q) 
= 12.95 kQ || 972.6 Q 

= 904.66 Q 

Z () = R ( = 2.2 kQ 



(b) - 






R l +z, 



_ 5.6 kQ(-226.2) = _ 16M 
“ 5.6 kQ + 2.2 kQ 



( d ) ^ = -<Y 



= -(-162.4) 



(904.66 Q) 
5.6 kQ 



www.elsolucionario.net 




(e) A, 



-R c \\R e _ -2.2 kQ 1 5.6 kQ 
r ~ 9.726 Q 



= -162.4 

Zi = 68 kQ || 16 kQ || 972.6 Q 



= 904.66 Q 

= (-162.4X904.66 Q) 
5.6 kQ 

= 26.24 

Z 0 = R C = 2.2 kQ 
Same results! 



(a) A, 



R r + Z n 

L o 



i? L = 4.7kQ: A 



R L = 2.2 kQ: A 



4.7 kQ 

4.7 kQ + 2.2 kQ 
2.2 kQ 

2.2 kQ + 2.2 kQ 
0.5 kQ 



(-226.4) = -154.2 



(-226.4) =-113.2 



R l = 0.5 kQ: A = (-226.4) = -41.93 

0.5 kQ + 2.2 kQ v 

A, X 



(b) Unaffected! 



v -V 

v CC v BE 



18 V-0.7 V 



R b + (J3 + 1)^ 680 kQ + (1 1 1)(0.82 kQ) 

= 22.44 /uA 

I E = (/?+ \)I B = (1 10 + 1)(22.44 //A) 

= 2.49 mA 
26 mV 26 mV 



I rr 2.49 mA 



10.44 Q 



VNL r e +R E 10.44 Q + 0.82 kQ 

= -3.61 

Zt = R b || Z b = 680 kQ || ( fir e + (/?+ 1)^) 

= 680 kQ || (610)(10.44 Q) + (1 10 + 1)(0.82 kQ) 
= 680 kQ || 92.17 kQ 
= 81.17 kQ 
Z 0 = R C = 3 kQ 



(b) - 



www.elsolucionario.net 




(c) A - V Q- A - 4J kQ (~ 3 * 61 ) 

VL V t R l +R q Vnl 4.7kQ + 3kQ 

= - 2.2 

, v n v n v. 

A L 

% V s V t V s 

y = j£_’ = 81.17 kQ (V s ) =0992 y 

1 Z { + R s 81.17 kQ + 0.6 kQ 
A Vs = (-2.2)(0.992) 

= - 2.18 



(d) None! 

(e) A Vl -none! 
V; Z, 



81.17 kQ 



= 0.988 



V s Z t + R s 81.17 kQ + lkQ 
A Vs = (-2.2)(0.988) 

= -2.17 

Rs't, A v i, (but only slightly for moderate changes in R s since Z z is typically much larger 
than R s ) 



43. Using the exact approach: 



F -V 

_ ^ Th y BE 

R Th + (fi + 1 )^ 

= 2.33 V- 0.7 V 

10.6 kQ + (121)(1.2 kQ) 
= 10.46 //A 



Erh ~ 



Jh v 

R x +R 2 cc 

12 kQ 



-(20 V) =2.33 V 



91 kQ + 12 kQ 
R Th = R l || R 2 = 91 kQ || 12 kQ = 10.6 kQ 



I E = 00+1)/* = (121)(10.46 juA) 
= 1.266 mA 
26 mV 26 mV 



r e = 



I E 1.266 mA 



= 20.54 Q 



(a) 



A 



V NL 



Re 

r e+ R E 



1.2 kQ 

20.54 Q + 1.2 kQ 



= 0.983 



Z; = R x II R 2 II (fir e + (J3+1)R e ) 

= 91 kQ || 12 kQ || ((120)(20.54 Q) + (120 + 1)(1.2 kQ)) 
= 10.6 kQ || (2.46 kQ + 145.2 kQ) 

= 10.6 kQ || 147.66 kQ 

= 9.89 kQ 



Z 0 = R e || r e = 1.2 kQ || 20.54 Q 

= 20.19 Q 



71 

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(b) - 

(c) A = 



R, 



R,+z, 



~A Nl 



= 0.976 

Z, 



A = 



Z.+/2„ 

= 0.92 



-4 = 



2.7 kQ(0.983) 

_ 2.7 kQ + 20.19 Q 

9.89 kQ(0.976) 
9.89 kQ + 0.6 kQ 



(d) A Vl =0.976 (unaffected by change in R s ) 
a _ z i a _ 9.89 kQ(0.976) 

V x ^7 , n Vr 



Z,. + fl 



9.89 kQ + 1 kQ 



= 0.886 (vs. 0.92 with R s = 0.6 kQ) 

As^t, A v i 

(e) Changing R s will have no effect on A v , Z h or Z Q . 



© < = 



R, 



R L +Z c 



-M= 



5.6 kQ(0.983) 



A = 



5.6 kQ + 20.19 Q 
= 0.979 (vs. 0.976 with R L = 2.7 kQ) 
9.89 kQ(0.979) 



-(Z v ) = - 

V V L J < 



Z . + R s " VL ' 9.89 kQ + 0.6 kQ 

= 0.923 (vs. 0.92 with R L = 2.7 kQ) 
As R l t, t, ^ t 



44. 



, Z £z? -Z gz? 6 V -0.7V 
(a) I E =— 



Re 

= 2.41 mA 



2.2 kQ 



26mV = J6m^ = ia79fl 
I E 2.41 mA 
= Rc_ = 4.7 kQ =435 S9 
r e 10.79 Q 

Z, = R e || r e = 2.2 kQ || 10.79 Q = 10.74 Q 
Z 0 = R C = 4.7 kQ 



(b) - 

W < = 

Vt = 



R^ a _ 5.6 kQ, 435.59, ^ 

v nl r ~ ~ - ~ - ~ 



Rl+R< 



Z, + ft 



-v = 



5.6 kQ + 4.7 kQ 
10.74 Q(ZJ 



10.74 Q + 100 Q 



= 0.097 V s 



A Vs =^--~ V = (236.83)(0.097) 

= 22.97 



72 

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(d) 



V, = I e -r e 
V 0 = -IoR l 

I _ -4.7 kQ(/J 

° 4.7 kQ + 5.6 kQ 



= -0.45634 



= K = +(O-450 e )R L 0-4563(5.6 kQ) 

1 V t /l e -r e ~ 10.79 Q 

= 236.82 (vs. 236.83 for part c) 



A Vs : 2.2 kQ || 10.79 Q = 10.74 Q 

Vi= V s = 10 - 74Q ( F ») =0.097 V s 

Z t +R s s 10.74 Cl + 100 Cl 

=(236.82X0.097) 

r i r s 

= 22.97 (same results) 



( e ) A £ = 



= 138.88 



’ A Vm 



2.2 kQ 



(435.59) 



A -HA H- 

% V, K ’ V Z, + R 



2.2 kQ + 4.7 kQ 
Z, 10.74 Q 



= 0.021 



10.74 Q + 500 Q 

Z v = (138.88)(0.021) = 2.92 

A v very sensitive to increase in R s due to relatively small Z z ; R s t, A v i 
A v sensitive to R L ; R L i, A v i 



(f) Z Q = R c = 4.7 kQ unaffected by value of R s \ 

(g) Zi = Re || r e = 10.74 Q unaffected by value of R L \ 



45. 



(a) 



\ = 



A = 



Rl+K 

r iA,, 



R, + R n 

L o 



lkQC-420) 
lkQ + 3.3 kQ 



2.7 kQ(-420) 
2.7 kQ + 3.3 kQ 



(b) A Vl = A Vi •A Vi = (-97.67X-189) = 18.46 x 10 3 

v V v v. 

Vs k n' v h ' v s 

W) , ft625 

Z . + 1 kQ + 0.6 kQ 

=(-189)(-97.67)(0.625) 

= 11.54 x 10 3 



73 

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^,-(-97.67)(lkQ) =9767 
1 R l 1 kfi 

A 2^ — (~ 1 89)(1 kQ) _ ?0 

' 2 /?, 2.7 kQ 

(d) 4 l = 4_ • 4 = (97.67X70) = 6.84 x 10 3 



(e) No effect! 

(f) No effect! 

(g) In phase 



46. 



(a) 4 = 



z. +z 

i 2 o 

= 0.984 

Rr 



A = 



Rr +Z n 



A Nr. 



~A, 



(i) 



1.2 kQ 

1 .2 kQ + 20 Q 
2.2 kQ 

2.2 kQ + 4.6 kQ 



(-640) 



= -207.06 



(b) A Vl =A Vi -A V2 =( 0.984)(-207.06) 

= -203.74 



4, =- 



4+4 Al 

50 kQ 

50 kQ + 1 kQ 

= -199.75 



(-203.74) 



(c) 



4 =~A 



z 

h_ 

z 

‘2 



= -(0.984) 



(50 kQ) 
1.2 kQ 



= -41 



4 = 




= -(-207.06) 



(1.2 kQ) 
2.2 kQ 



= 112.94 



(d) 



A=~A 



z 



= -(-203.74) 



(50 kQ) 
2.2 kQ 



= 4.63 x 10 3 



74 

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(e) A load on an emitter-follower configuration will contribute to the emitter resistance (in 
fact, lower the value) and therefore affect Z { (reduce its magnitude). 

(f) The fact that the second stage is a CE amplifier will isolate Z Q from the first stage and R s . 

(g) The emitter-follower has zero phase shift while the common-emitter amplifier has a 
180° phase shift. The system, therefore, has a total phase shift of 180° as noted by the 
negative sign in front of the gain for A Vt in part b. 



47. For each stage: 

_ 6.2 kQ 

V B - 



(15 V) = 3.08 V 



24 kQ + 6.2 kQ 
V E =V B - 0.7 V = 3.08 V - 0.7 V = 2.38 V 

r r V E 2.38 V 1 A 

I F = I c = — = = 1.59 mA 

R e 1.5 kQ 

V c = V cc -IcRc= 15 V - (1.59 mA)(5.1 kQ) 

= 6.89 V 



48. 



26 mV 26 mV 



I E 1.59 mA 



= 16.35 Q 



R, = 



= Ri || R 2 1| pr e = 6.2 kQ || 24 kQ || (150)(16.35 Q) 
= 1.64 kQ 



A = — 



RrWR, 5.1 kQ || 1.64 kQ 



= -75.8 



A ~- 

v 2 

A v = 



16.35 Q 

= _ 3119 

r e 16.35 Q 

A v A Vi = (—7 5 . 8)(— 3 11.9) = 23,642 



49. 



V a =- 



3.9 kQ 



K = 



3.9 kQ + 6.2 kQ + 7.5 kQ 
6.2 kQ + 3.9 kQ 
3.9 kQ + 6.2 kQ + 7.5 kQ 



(20 V) = 4.4 V 
(20 V) =11.48 V 



= V B - 0.7 V = 4.4 V - 0.7 V = 3.7 V 



Ir =h =^ = 



\ _ 3.7 V _ 



R e 1 kQ 



= 3.7 mA = A =I r 



= V cc ~ IcRc = 20 V - (3.7 mA)(l .5 kQ) 

= 14.45 V 



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50 . 



26 mV 26 mV 

r e = = =7fi 

I E 3.7 mA 



r , 

t R e 1.5 kn ... 

A =-^ = =214 

2 ^ 7 Q 

+ = + + = (-1X214) = -214 
V 0 = A v V i = (-214)(10 mV) = -2.14 V 



51. R 0 = R d = 1-5 kQ (V 0 (from problem 50) = -2.14 V) 

V 0 (load) = (V) = —^—(-2.14 V) 

R o + R E 1 0 kO + 1.5 kQ 

= -1.86 V 



52. 



V -V 

v rr v m 



(16 V-1.6 V) 



J3 d R e + R b (6000)(510 Q) + 2.4 MQ 

14.4 V . 

= = 2.64 uA 

5.46 MQ 

I c = h = PeJb = 6000(2.64 juA) = 15.8 mA 
V E = I e Re = (15.8 mA)(510 Q) = 8.06 V 



53. From problem 69, I E = 15.8 mA 

26 26 V i r, 

r P = — = = 1 .65 Q 



A v = 



I E 15.8 mA 
R , 



510 Q 



r 0 + R e 1.65 Q + 510 Q 



= 0.997 * 1 



54. 



v -V 

dc: h= cc BE — 



16 V-1.6V 



r b + Pd r e 2A MQ + (6000)(510 Q) 

I c = p D I B = (6000)(2.64 juA) = 15.84 mA 

26 mV 26 mV t _ 

r = = = 1.64 Q 

62 I E 15.84 mA 



= 2.64 juA 



ac: Z z = p D r ei = (6000)(1 .64 Q) = 9.84 kQ 

V, 



9.84 kQ 



V 0 = (~P D I b p(R c ) = -(6000) 



= -121.95 Vi 



V 



9.84 kQ 



(200 Q) 



and A =-^= -121.95 
V, 



16 

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55 . 



= Vcc~ V eb x = 16 V-0.7 V 

5 R b + p x p 2 R E "1.5MQ + (160)(200)(100 Q) 

= 3.255 juA 

I c = pip2l B = ( 1 60)(200)(3 .255 //A) = 104.2 mA 

V Ci = Vcc-IcRc= 16 V- (104.2 mA)(100Q) = 5.58 V 

V Bi = I b R b = (3.255 //A)(1.5 MQ) = 4.48 V 



56. From problem 55: I E = 0.521mA 

26 mV 26 mV _ 

r = = = 49.9 Q 

1 I E (mA) 0.521 mA 

R. = pr e = 160(49.9 Q) = 7.98 kQ 

A _ PARc _ (160)(200)(100Q) 

v P\P 2 R c + \ (160)(200)(100 Q) + 7.98 kQ 

= 0.9925 

V 0 =A v Vi =0.9975 (120 mV) 

= 119.7 mV 



57. 



26 mV 26 mV 



^E(dc) ^ *2 m A 



= 21.67 Q 



pr e = (120)(21.67 Q) = 2.6 kQ 



58. 



59. 

60. 
61. 



62. 



(a) = — = -160 

Vi 

V„ = -160 Vi 



(b) 



j _ v-Kv 0 v-KW m -MJ 

6 K K h ie 

FJ (l - (2 x 10 -4 )(1 60)) 

I kQ 

l b = 9.68 x 10 ~ 4 V t 



(c) 




= 1 x 10 - 3 V, 



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(d) % Difference = 



1 x 1 0 K 



x 100% 



(e) Valid first approximation 



% difference in total load = 



Rl-RlPk 



X 100% 



2.2 kQ-(2.2kQ||50kQ) 
2.2 kQ 

2.2 kQ- 2.1073 kQ _ 



x 100% 



2.2 kQ 



x 100% 



In this case the effect of 1 !h oe can be ignored. 

(a) Vo = -180 Vi (h ie = 4 kQ, h re = 4.05 x 1 0“ 4 ) 
n,, , _ ^ - (4.05 xl(T 4 )(l 8019 



= 2.32 x lO^F, 

V V A 

(c) Ib= 7^ = JT^ =2 - 5x10 ' V ‘ 

h. 4 kQ 



(d) % Difference : 



2.5x10 if -2.32x10 Jf 
2.5x10 4 K 



x 100% = 7.2% 



(e) Yes, less than 10% 
From Fig. 5.18 



mm max 



h oe : 1 /uS 30 /uS 



(1 + 30)//S 

Avg = f 2~~ = 15 ’ 5 ^ 



(a) h fe = JS= 120 

= pr e = (120)(4.5 Q) = 540 Q 

h oe = — = — - — = 25 //S 
r 40 kQ 



(b) r,^=‘“=lLUfl 
p 90 



P=h fe = 90 



Ke 20 jLlS 

50 kQ 



www.elsolucionario.net 




67. (a) r e = 8.31 Q (from problem 9) 

(b) h fe = P= 60 

h ie = pr e = (60)(8.31 Q) = 498.6 Q 

(c) Z, = Rg II h ie = 220 kQ || 498.6 Q = 497.47 Q 
Z 0 = R C = 2.2 kQ 



,d) 



K 

Ai = hf e = 60 



498.6 Q 



(e) Zi = 497.47 Q (the same) 

= r a || R c , r„ = — 

25 /uS 

= 40 kQ || 2.2 kQ 

= 2.09 kQ 



= 40 kQ 



-y.K> = -(60X2.085 kQ). 

^ 498.6 Q 

At = -A v ZJRc = -(-250.90)(497.47 Q)/2.2 kQ = 56.73 

68. (a) 68 kQ || 12 kQ = 10.2 kQ 

Zi = 10.2 kQ || = 10.2 kQ || 2.75 kQ 

= 2.166 kQ 
Z 0 — Rc II r o 

= 2.2 kQ || 40 kQ 
= 2.085 kQ 



~ h fe R C 

(b) 

h 



R'c =R C \\ r 0 = 2.085 kQ 



~(180)(2.085 kQ) 
2.75 kQ 



= -136.5 



Ai =L = LA 

L L 



h 



l l 



1 + Ke R L J 



10.2 kQ 



10.2 kQ + 2.68 kQ 



180 



1 + (25 //S)(2.2 kQ) 

= 135.13 



(0.792) 



79 

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69 . 



(a) Zi-R E \\hib 

= 1.2 kQ || 9.45 Q 

= 9.38 Q 

Z 0 = R C || — = 2.7 kQ || - — - = 2.7 kQ || 1 MQ = 2.7 kQ 

h <* lxlO ” 6 — 

V 

(b) A _ -h^Rcp/Kh) = -(-0.992)(= 2.7 kQ) 

1 J v h ib 9.45 Q 

= 284.43 

4 = -l 



(c) 



a=-hj b = -(-0.992) = 0.992 



or _ 0.992 
1-a" 1-0.992 



124 



r e = h ib = 9.45 Q 

1 1 

r °~K~ b ~ 1//A/V 



1 MQ 



70. 



(a) Z/=^- 



h feKe R L 
l + Ke R L 



= 2.15 kQ- 



(180)(2xl0" 4 )(2.2kQ) 
(l + 25//S)(2.2 kQ) 



= 2.68 kQ 



Zi = 10.2 kQ || Z t ! = 2.12 kQ 

Z '= 1 - I 

° h oe -(h fe h re /K) 25 //S-(180)(2x 10” 4 )/2.75 kQ 
= 83.75 kQ 



Z 0 = 2.2 kQ || 83.75 kQ = 2.14 kQ 



-h fe R L — (180)(2.2 kQ) 

” h ie+( h ie h oe- h feKe) R L ~ 2.75 kQ + ((2.75 kQ)(25//S) - (1 80)(2 X 1 0" 4 )) 2.2 kQ 

= -140.3 



(c) 



A = - 



n fe 



1 + hR r 



( 180 ) 

l + (25//S)(2.2 kQ) 



= 170.62 




I I 

— — — = (170.62) 
i! i t 



10.2 kQ ^ 
10.2 kQ + 2.68 kQ > 



= 135.13 



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71 . 



7 — h — hf e h r eR L 

(a) Zi hi e 

1 + Ke R L 



= 0.86 kQ 



(140)(1.5xl0” 4 )(2.2 kQ) 



1 + (25 //S)(2.2kQ) 

= 0.86 kQ- 43.79 Q 
= 816.21 Q 

= R B \\Zi 

= 470 kQ || 816.21 Q 

= 814.8 Q 



(b) A v = 



~ h fe R L 



h ie+( h ieKe- h feKe) R L 

-(140)(2.2 kQ) 



0.86 kQ + ((0.86 kQ)(25 //S)-(140)(1.5xl0" 4 ))2.2 kQ 

= -357.68 



(c) Ai=^ = 



h 



' fe 



140 



/. 1 + h oe R L 1 + (25 //S)(2.2 kQ) 

132.70 



4 =-7 = 



(/ 1 


/ \ 

/. 




/_ 


UJ 


U'J 



/,= 



470 kQ I 



(d) Z 0 = 



= (132.70X0.998) 

= 132.43 
1 



470 kQ + 0.816 kQ 

^ = 0.998 
/' 



72. (a) Z,. 



Z,= 

(b) 4' = 

4 = 



/* oe -{hfAeKK + *,)) 25 //S -((140)(1 .5/10 4 )/(0.86 kQ + 1 kQ)) 

= l - = 72.9 kQ 

13.71 //S 

(~0.997)(1 x!Q- 4 )(2.2kQ) 

1 + (0.5 i «A/V)(2.2 kQ) 

= 9.67 Q 

1.2 kQ || Z,' = 1.2 kQ || 9.67 Q = 9.59 Q 



h n h,R r 

, ft, rb - 9- 45 Q — 

l+« 



-0.997 



1 + 1 + (0.5 //A/V)(2.2 kQ) 

— - — = (-0.996) f L2k ° 

/' I . \l.2 kQ + 9.67 kQ 



- = -0.996 



= -0.988 



81 

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h ,b+(KKb- h fl,Kb) R L 

-(-0.997)(2.2 kQ) 

~ 9.45 Q + ((9.45 Q)(0.5 i «A/V)-(-0.997)(lxl0 4 ))(2.2 kD) 

= 226.61 



° Kb-[hjbKblh lb ] 

1 

~ 0.5 juA/Y - [(-0.997)(1 x 1 0” 4 ) / 9.45 o] 
= 90.5 kQ 

Z Q = 2.2 kQ || Z' = 2.15 kQ 



73. 

74. (a) h fe (0.2 mA) = 0.6 (normalized) 

h fe (1 mA) =1.0 



% change 



h fe (0.2 mA) ~h fe (l mA) 
h fe (0.2 mA) 



x 100% 



= 0 . 6-1 
0.6 

= 66.7% 



x 100% 



(b) 



h fe { 1 mA) =1.0 
h fe { 5 mA) =1.5 



% change 



/z /e (lmA)-/z /e (5mA) 

^(ImA) 



x 100% 



1-1.5 

1 



x 100% 



= 50% 



75. Log-log scale! 



(a) I c = 0.2 mA, = 4 (normalized) 
4=1 mA, h ie = 1 (normalized) 

1 4 — 1 1 



% change = 



x 100% = 75% 



(b) 4 = 5 mA, = 0.3 (normalized) 
Il-0.3| 



% change = 



1 



x 100% = 70% 



82 

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76. (a) h oe = 20 //S @ 1 mA 

I c = 0.2 mA, h oe = 0.2(h oe @ 1 mA) 

= 0 . 2(20 juS) 

= 4 juS 

(b) r Q = — = — — =250kQ » 6.8 kQ 
Ke 4 MS 
Ignore i/h oe 

77. (a) I c = 10 mA, h oe = 10(20 //S) = 200 //S 

(b) r 0 = — = — l - — =5kQvs. 8.6 kQ 
h oe 200//S 

Not a good approximation 

78. (a) Kl 0. 1 mA) = 4{h re (\ mA)) 

= 4(2 x 10 -4 ) 

= 8 x 10‘ 4 

(b) h re V ce = h re A v ■ Vj 

= (8 x 1 0 _4 )(2 1 0) F, 

= 0.168 V t 

In this case h re V ce is too large a factor to be ignored. 

79. (a) h fe 

(b) h oe 

(c) h oe = 30 (normalized) to 

h oe = 0.1 (normalized) at low levels of I c 

(d) mid-region 

80. (a) h ie is the most temperature-sensitive parameter of Fig. 5.33. 

(b) h oe exhibited the smallest change. 

(c) Normalized: h Mmax) = 1.5, h fe(min) = 0.5 

For hf e =100 the range would extend from 50 to 150 — certainly significant. 

(d) On a normalized basis r e increased from 0.3 at -65°C to 3 at 200°C — a significant 
change. 

(e) The parameters show the least change in the region 0° — » 100°C. 



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81. 



(a) Test: 



J3R e > 10R 2 

70(1.5 kQ)> 10(39 kQ) 
? 

105 kQ > 390 kQ 
No! 



R rh = 39 kQ || 150 kQ = 30.95 kQ 
39 kQ(14 V) 






Ib = 



39 kQ + 150 kQ 
E -V 

^ Th v BE 



= 2.89 V 



2.89 V- 0.7 V 



R n + (/? + 1 )R e 30.95 kQ + (71)(1.5 kQ) 



= 15.93 juA 
Vb = E T h ~ hRn 

= 2.89 V - (15.93 //A)(30.95 kQ) 
= 2.397 V 



Ve = 2.397 V - 0.7 V = 1.697 V 



and I E = — 

E e 



1.697 V 
1.5 kQ 



= 1.13 mA 



Vce — Vcc — Ic(Rc + Re) 

= 14 V - 1.13 mA(2.2 kQ + 1.5 kQ) 

= 9.819 V 

Biasing OK 

(b) R 2 not connected at base: 
j _ V cc -0 _ 14 V-0.7 V 

B R b +(P + \)R e 150kQ + (71)(1.5kQ) 

V B = V cc -IbRb= 14 V-(51.85 /iA)(150 kQ) 
= 6.22 V as noted in Fig. 5.187. 



51.85 juA 



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Chapter 6 



WWW. 



233.33 Q 
0.5625 



933.2 Q 




(c) V GS = -2V,I D = 2mA 
V gs = -3 V, I D = 0.5 mA 

A I D = 1.5 mA 

(d) F GS = -3V,/ fl = 0.5mA 
V gs = -4 V,I d = 0mA 

A I D = 0.5 mA 

(e) As V GS becomes more negative, the change in I D gets progressively smaller for the same 
change in V GS . 

(f) Non-linear. Even though the change in V GS is fixed at 1 V, the change in I D drops from a 
maximum of 3.5 mA to a minimum of 0.5 mA — a 7:1 change in A I D . 

5. The collector characteristics of a BJT transistor are a plot of output current versus the output 
voltage for different levels of input current. The drain characteristics of a JFET transistor are 
a plot of the output current versus input voltage. For the BJT transistor increasing levels of 
input current result in increasing levels of output current. For JFETs, increasing magnitudes 
of input voltage result in lower levels of output current. The spacing between curves for a 
BJT are sufficiently similar to permit the use of a single beta (on an approximate basis) to 
represent the device for the dc and ac analysis. For JFETs, however, the spacing between the 
curves changes quite dramatically with increasing levels of input voltage requiring the use of 
Shockley’s equation to define the relationship between I D and V GS . V c and V P define the 

region of nonlinearity for each device. 

6. (a) The input current I G for a JFET is effectively zero since the JFET gate-source junction is 

reverse-biased for linear operation, and a reverse-biased junction has a very high resistance. 

(b) The input impedance of the JFET is high due to the reverse-biased junction between the 
gate and source. 

(c) The terminology is appropriate since it is the electric field established by the applied 
gate to source voltage that controls the level of drain current. The term “field” is 
appropriate due to the absence of a conductive path between gate and source (or drain). 

7. V GS = 0 V, To = I DSS = 12 mA 
V GS = V P = -6 V,/d = 0mA 

Shockley’s equation: V GS = -1 V, I D = 8.33 mA; V GS = -2 V, I D = 5.33 mA; V GS = -3 V, 

I D = 3 mA; V GS = -4 V, I D = 1.33 mA; V GS = -5 V, I D = 0.333 mA. 




8. For a ^-channel JFET, all the voltage polarities in the network are reversed as compared to an 
^-channel device. In addition, the drain current has reversed direction. 



86 

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9. 

10 . 



11 . 



12 . 



13. 



14. 



15. 



(b) Idss = 10 mA, V P = -6\ 

Vgs = 0 V, //) = loss = 12 mA 
V G s= V P = - 4 V, 7p> = 0 mA 

Kgs = ^ = -2 V, I D = %=3mA 
2 4 

F G5 = 0.3 = -1.2 V, A) = 6 mA 

F G 5 = -3 V, /d = 0.75 mA (Shockley’s equation) 

(a) Id = Idss = 9 mA 

(b) /d = Idss( 1 ~ VgsIVp) 2 

= 9 mA(l - (-2 V)/(-3.5 V)) 2 

= 1.653 mA 

(c) V GS = Vp = -3.5 V, ip = 0 mA 

(d) F G5 < Fp = -3.5 V, ip = 0 mA 
F G * = 0 V,7 D =16mA 

F G5 = 0.3 V P = 0.3(-5 V) = -1.5 V, I D = W2 = 8 mA 
F G * = 0.5 Fp = 0.5(-5 V) = -2.5 V, I D = W4 = 4 mA 

F G <? = Fp = -5 V, 7p> = 0 mA 



F Gl s - 0 V, Id - Idss - 7.5 mA 

V GS = 0.3 V P = (0.3)(4 V) = 1 .2 V, I D = Idss/2 = 7.5 mA/2 = 3.75 mA 
V GS = 0.5 V P = (0.5)(4 V) — 2 V, I D = W4 = 7.5 mA/4 = 1.875 mA 
Vgs = Fp = 4 V, ip = 0 mA 



(a) 7 D = 7^(1 - VgsIVp) 2 = 6 mA(l - (-2 V)/(-4.5 V)) 2 

= 1.852 mA 

Id = 7^(1 - VgsIVp) 2 = 6 mA(l - (-3.6 V)/(-4.5 V)) 2 



= 0.24 mA 

3 mA 
6 mA 

V , — y V / 

= -1.318 V 

5.5 mA " 
6 mA 

= -0.192 V 

7 p > = Idss( 1 ~ VgsIVp) 2 
3 mA = /nssO - (-3 V)/(-6 V)) 2 
3 mA = W0.25) 



V G s= V d 



l-' 1 * 



= (-4.5 V) 



1 - 



( 



(b) V GS =V P 



,-,A. 

1 n w 



= (-4.5 V) 



1- 



Idss — 12 mA 



87 



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18 . 



V GS = -0.5 V,I D = 6.5 mA . . 

y. , r . A r 2.5 mA 
V GS = -1 V, I D = 4 mA 

Determine A/ D above 4 mA line: 

2.5 mA x , 

= => x = 1 .5 mA 

0.5 V 0.3 V 

I D = 4 mA +1.5 mA = 5.5 mA corresponding with values determined from a purely 
graphical approach. 



19. 

20 . 



22 . 



Yes, all knees of V GS curves at or below \ V P \ = 3 V. 

From Fig 6.25, I DS s = 9 mA 
At V GS = - 1 V,/z) = 4mA 
Id = Idss( 1 _ Vcs/Vp) 2 

D ~ = 1 - Vgs/Vp 



V p 



= 1 - 



V P = 



-IV 



i- A 

I nee 



1 - 



4 mA 
9 mA 



= -3 V (an exact match) 

21. / D = W1 - IWKp) 2 

= 9 mA(l - (-1 V)/(— 3 V)) 2 

= 4 mA, which compares very well with the level obtained using Fig. 6.25. 



(a) V D s = 0.7 V @ A) = 4 mA (for V GS = 0 V) 



A/ n 



0.7V-0V 



4 mA- 

(b) For Vgs = -0.5 V, ( 
0.7 V 



= 175 Q 



3 mA 



= 233 Q 



(c) 'V = 



0mA 

} /d = 3 mA, Vds = 0.7 V 
175 Q 



(1-F^/Fy,) 2 (1 — (—0.5 V)/(-3 V) 2 



= 252 Q vs. 233 Q from part (b) 



23. 



24. The construction of a depletion-type MOSFET and an enchancement-type MOSFET are 

identical except for the doping in the channel region. In the depletion MOSFET the channel is 
established by the doping process and exists with no gate-to-source voltage applied. As the 
gate-to-source voltage increases in magnitude the channel decreases in size until pinch-off 
occurs. The enhancement MOSFET does not have a channel established by the doping 
sequence but relies on the gate-to-source voltage to create a channel. The larger the 
magnitude of the applied gate-to-source voltage, the larger the available channel. 



89 



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25 . 



26. At V GS = 0 V, / D = 6 mA 

At V GS = -l V, In = 6 mA(l - (-1 V)/(-3 V)) 2 = 2.66 mA 

At F gs = +1 V, I D = 6 mA(l - (+1 V)/(-3 V)) 2 = 6 mA(l ,333) 2 = 10.667 mA 

At V GS = +2 V, I D = 6 mA(l - (+2 V)/(-3 V)) 2 = 6 mA(1.667) 2 = 16.67 mA 

Vgs Id 

-1 V 2.66 mA 
0 6.0 mA 

+1 V 10.67 mA 
+2 V 16.67 mA 

From -1 V to 0 V, M D = 3.34 mA 

while from +1 V to +2 V, AI D = 6 mA - almost a 2: 1 margin. 

In fact, as V GS becomes more and more positive, I D will increase at a faster and faster 
rate due to the squared term in Shockley’s equation. 



A I D = 3.34 mA 



} 



A I D = 4.67 mA 



A I D = 6 m A 



27. 



Vgs = 0 V, l D = I DSS = 12 mA; V GS = -8 V, I D = 0 mA; V GS = A = ~ 4 V, = 3 mA; 

2 

V GS = 0.3 V P = -2.4 V, I D = 6 mA; = -6 V, = 0.75 mA 



28. 



From problem 20: 




+1 V 



1- 



14 mA 
9.5 mA 



1 

-0.21395 



= -4.67 V 



+1 V _ +1 V 

1 — Vl .473 ~ 1-1.21395 



29. 



Id = Idss( 1 - V GS IVp) 2 



loss - 



(i-v GS /v P ) 



4 mA 

(l-(-2V)/(-5V)) 2 



= 11.11 mA 



30. From problem 14(b" 


): 






i 


i 


I D 1 




f 1 20 mA 


ii 

il 


1-J 




= (-5 V) 






l V 


I DSS J 




[ V 2.9 mA J 



= (-5 V)(l - 2.626) = (-5 VX-1.626) 

= 8.13 V 



31. 



From Fig. 6.34, P D = 200 mW, I D = 8 mA 



P=V DS I D 

and V DS = — 
In 



200 mW 
8 mA 



= 25 V 



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32. (a) In a depletion-type MOSFET the channel exists in the device and the applied voltage 

V GS controls the size of the channel. In an enhancement-type MOSFET the channel is 
not established by the construction pattern but induced by the applied control voltage 

Vgs • 

(b) - 

(c) Briefly, an applied gate-to-source voltage greater than V T will establish a channel 
between drain and source for the flow of charge in the output circuit. 

33 . (a) I D = k( V GS ~ V T f = 0.4 x 1 0" 3 ( V GS - 3 ,5) 2 



(b) 



Vgs Id_ 



3.5 V 


0 


4 V 


0.1 mA 


5 V 


0.9 mA 


6 V 


2.5 mA 


7 V 


4.9 mA 


8 V 


8.1 mA 


II 

p 

oo 

X 

o 


-\V GS - 3.5) 2 


Vgs 


Id 



3.5 V 


0 


For same levels of V GS , Id attains 


4 V 


0.2 mA 


twice the current level as part (a). 


5 V 


1.8 mA 


Transfer curve has steeper slope. 


6 V 


5.0 mA 


For both curves, I D = 0 mA for 


7 V 


9.8 mA 


V GS <3.5 V. 


8 V 


16.2 mA 





34. 



(a) k = 



D(on) 



4mA 



(V -V Y (6 V - 4 V) 2 

v GS(on) 'T ) 



= 1 mA/V 2 



Id = Was ~ Vt ) 2 = 1 x USVas ~ 4 V) 2 



(b) 



(c) 



V GS 


Id 


4 V 


0 mA 


5 V 


1 mA 


6 V 


4 mA 


7 V 


9mA 


8 V 


16 mA 


Vgs 


Id 


2 V 


OmA 


5 V 


1 mA 


10 V 


36 mA 



For Vgs < Vt = 4V,/ n = 0 mA 



(V GS < V t ) 



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